Aptitude & reasoning

 

 

 

Introduction:

There are four main directions – East, West, North and South as shown below:

 

 

 

 

There are four cardinal directions – North-East (N-E), North-West (N-W), South-East (S-E), and South-West (S-W) as shown below:

 

 

 

Key points

 

1. At the time of sunrise if a man stands facing the east, his shadow will be towards west.
2. At the time of sunset the shadow of an object is always in the east.
3. If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.
4. At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow

 

 

 

 

 

 

 

 

 

Practice Questions

Type 1:

Siva starting from his house, goes 5 km in the East, then he turns to his left and goes 4 km. Finally he turns to his left and goes 5 km. Now how far is he from his house and in what direction?

Solution:

From third position it is clear he is 4 km from his house and is in North direction.

 

 

 

 

 

 

Type 2:

Suresh starting from his house, goes 4 km in the East, then he turns to his right and goes 3 km. What minimum distance will be covered by him to come back to his house?

Solution:

 

Type 3:

One morning after sunrise Juhi while going to school met Lalli at Boring road crossing. Lalli’s shadow was exactly to the right of Juhi. If they were face to face, which direction was Juhi facing?

 

Solution: In the morning sunrises in the east.

So in morning the shadow falls towards the west.

Now Lalli’s shadow falls to the right of the Juhi. Hence Juhi is facing South.

 

 

 

 

Type 4: Hema starting from her house walked 5 km to reach the crossing of Palace. In which direction she was going, a road opposite to this direction goes to Hospital. The road to the right goes to station. If the road which goes to station is just opposite to the road which IT-Park, then in which direction to Hema is the road which goes to IT-Park?

Solution:

From II it is clear that the road which goes to IT-Park is left to Hema.

 

 

 

 

 

 

 

 

Questions

 

Level-1

 

1.	One morning Udai and Vishal were talking to each other face to face at a crossing. If Vishal’s shadow was exactly to the left of Udai, which direction was Udai facing?
	A. East B. West C. North D. South

2.	Y is in the East of X which is in the North of Z. If P is in the South of Z, then in which direction of Y, is P?
	A. North B. South C. South-East D. None of these

3.	If South-East becomes North, North-East becomes West and so on. What will West become?
	A. North-East B. North-West C. South-East D. South-West

4.	A man walks 5 km toward south and then turns to the right. After walking 3 km he turns to the left and walks 5 km. Now in which direction is he from the starting place?
	A. West B. South C. North-East D. South-West

5.	Rahul put his timepiece on the table in such a way that at 6 P.M. hour hand points to North. In which direction the minute hand will point at 9.15 P.M. ?
	A. South-East   B. South   C. North D. West
6.	Rasik walked 20 m towards north. Then he turned right and walks 30 m. Then he turns right and walks 35 m. Then he turns left and walks 15 m. Finally he turns left and walks 15 m. In which direction and how many metres is he from the starting position?
	A. 15 m West B. 30 m East C. 30 m West D. 45 m East

7.

Two cars start from the opposite places of a main road, 150 km apart. First car runs for 25 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point?

A. 65 km B. 75 km C. 80 km D. 85 km

8.	Starting from the point X, Jayant walked 15 m towards west. He turned left and walked 20 m. He then turned left and walked 15 m. After this he turned to his right and walked 12 m. How far and in which directions is now Jayant from X?
	A. 32 m, South B. 47 m, East C. 42 m, North D. 27 m, South

 

9.	One evening before sunset Rekha and Hema were talking to each other face to face. If Hema’s shadow was exactly to the right of Hema, which direction was Rekha facing?
	A. North B. South C. East D. Data is inadequate

10.	A boy rode his bicycle Northward, then turned left and rode 1 km and again turned left and rode 2 km. He found himself 1 km west of his starting point. How far did he ride northward initially?
	A. 1 km B. 2 km C. 3 km D. 5 km

 

 

Answers:

1Answer: Option C

Explanation:

 

2Answer: Option D

Explanation:

P is in South-West of Y.

 

3Answer: Option C

Explanation:

It is clear from the diagrams that new name of West will become South-East.

 

4Answer: Option D

Explanation:

Hence required direction is South-West.

 

5Answer: Option D

Explanation:

At 9.15 P.M., the minute hand will point towards west.

 

6Answer: Option D

Explanation:

 

7Answer: Option A

Explanation:

 

 

 

 

8Answer: Option A

Explanation:

 

9Answer: Option B

Explanation:

In the evening sun sets in West. Hence then any shadow falls in the East. Since Hema’s shadow was to the right of Hema. Hence Rekha was facing towards South.

 

10Answer: Option B

Explanation:

The boy rode 2 km. Northward

 

 

Level – 2

 

Dev, Kumar, Nilesh, Ankur and Pintu are standing facing to the North in a playground such as given below:

1. Kumar is at 40 m to the right of Ankur.
2. Dev is are 60 m in the south of Kumar.
3. Nilesh is at a distance of 25 m in the west of Ankur.
4. Pintu is at a distance of 90 m in the North of Dev

 

 

1.	Which one is in the North-East of the person who is to the left of Kumar?
	A. Dev B. Nilesh C. Ankur D. Pintu

2.	If a boy starting from Nilesh, met to Ankur and then to Kumar and after this he to Dev and then to Pintu and whole the time he walked in a straight line, then how much total distance did he cover?
	A. 215 m B. 155 m C. 245 m D.   185 m

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. Six flats on a floor in two rows facing North and South are allotted to P, Q, R, S, T and U.
2. Q gets a North facing flat and is not next to S.
3. S and U get diagonally opposite flats.
4. R next to U, gets a south facing flat and T gets North facing flat.

 

 

3.	If the flats of P and T are interchanged then whose flat will be next to that of U?
	A. P B. Q C. R D. T

4.	Which of the following combination get south facing flats?
	A. QTS B. UPT C. URP D. Data is inadequate

5.	The flats of which of the other pair than SU, is diagonally opposite to each other?
	A. QP B. QR C. PT D. TS
6.	Whose flat is between Q and S?
	A. T B. U C. R D. P

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. 8-trees → mango, guava, papaya, pomegranate, lemon, banana, raspberry and apple are in two rows 4 in each facing North and South.
2. Lemon is between mango and apple but just opposite to guava.
3. Banana is at one end of a line and is just next in the right of guava or either banana tree is just after guava tree.
4. Raspberry tree which at one end of a line, is just diagonally opposite to mango tree.

 

 

	7 .Which of the following statements is definitely true?
	A. Papaya tree is just near to apple tree. B. Apple tree is just next to lemon tree. C. Raspberry tree is either left to Pomegranate or after. D. Pomegranate tree is diagonally opposite to banana tree.

8	Which tree is just opposite to raspberry tree?
	A. Papaya B. Pomegranate C. Papaya or Pomegranate D. Data is inadequate
9	Which tree is just opposite to banana tree?
	A. Mango B. Pomegranate C. Papaya D. Data is inadequate

 

 

Answer: 1 Option D

Explanation:

Ankur is in the left of Kumar. Hence Pintu is in North-East of Ankur

 

 

Answer: 2 Option A

Explanation:

Required distance = 25 m + 40 m + 60 m + 90 m

Required distance = 215 m

 

 

Answer:3 Option C

Explanation:

Hence flat R will be next to U.

 

 

Answer:4 Option C

 

Explanation:

Hence URP flat combination get south facing flats.

 

Answer:5 Option A

 

Explanation:

Hence QP is diagonally opposite to each other.

 

 

 

 

 

 

Answer:6 Option A

 

Explanation:

Hence flat T is between Q and S.

 

Answer: 7 Option B

 

Explanation:

 

 

Answer:8 Option C

 

Explanation:

 

 

 

 

 

 

 

Answer:9 Option A

 

Explanation:

 

In this module we will deal with basic concepts of time and distance, speed, average speed, conversion from km/h to m/s and vice versa. This chapter will form the basis of further concept of relative speed which is used in train and boat problems.

Important Formulas

1. Speed=Distance/Time
2. Distance=Speed×Time
3. Time=Distance/Speed
4. To convert Kilometers per Hour(km/hr) to Meters per Second(m/s)
   x km/hr=(x×5)/18m/s
5. To convert Meters per Second(m/s) to Kilometers per Hour(km/hr)
   x m/s=(x×18)/5 km/hr
6. If a car covers a certain distance at x kmph and an equal distance at y kmph, the average speed of the whole journey = 2xy/(x+y) kmph
7. Speed and time are inversely proportional (when distance is constant) ⇒Speed ∝1/Time (when distance is constant)
8. If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is 1/a:1/b or b : a

Solved Examples

Level 1

1.A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
A. 8.2	B. 4.2
C. 6.1	D. 7.2

 

Answer : Option D

Explanation :

Distance = 600 meter

time = 5 minutes = 5 x 60 seconds = 300 seconds

Speed = distance/time=600/300=2m/s=(2×18)/5 km/hr=36/5 km/hr=7.2 km/hr

2.Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?
A. 17 hr	B. 14 hr
C. 12 hr	D. 19 hr

 

Answer : Option A

Explanation :

Relative speed = 5.5 – 5 = .5 kmph (because they walk in the same direction)

distance = 8.5 km

Time = distance/speed=8.5/.5=17 hr.

3.Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?
A. 1 hr 42 min	B. 1 hr
C. 2 hr	D. 1 hr 12 min

 

Answer : Option D

Explanation :

New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time – usual time = 12 minutes
=>1/6 of usual time = 12 minutes => usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes

4.A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?
A. 3 km	B. 4 km
C. 5 km	D. 6 km

 

Answer : Option D

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph,the average speed of the whole journey = 2xy/(x+y) kmph

Hence, average speed = (2×3×2)/(2+3)=12/5 km/hr .

Total time taken = 5 hours

⇒Distance travelled=(12/5)×5=12 km

⇒Distance between his house and office =12/2=6 km

5.If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. What is the actual distance travelled by him?
A. 80 km	B. 70 km
C. 60 km	D. 50 km

 

Answer : Option D

Explanation :

Assume that the person would have covered x km if travelled at 10 km/hr

⇒Speed = Distance/Time=x/10….. (Equation1)

Give that the person would have covered (x + 20) km if travelled at 14 km/hr
⇒Speed = Distance/Time=(x+20)/14….. (Equation2)

From Equations 1 and 2,
X/10=(x+20)/14⇒14x=10x+200⇒4x=200⇒x=200/4=50

6.A car travels at an average of 50 miles per hour for 212 hours and then travels at a speed of 70 miles per hour for 112 hours. How far did the car travel in the entire 4 hours?
A. 210 miles	B. 230 miles
C. 250 miles	D. 260 miles

 

Answer : Option B

Explanation :

Speed1 = 50 miles/hour

Time1 = 2*(1/2) hour=5/2 hour

⇒Distance1 = Speed1 × Time1 = (50×5)/2=25×5=125 miles

⇒Speed2 = 70 miles/hour

Time2 = 1*1/2 hour=3/2 hour

Distance2 = Speed2 × Time2 = 70×3/2=35×3=105 miles

Total Distance = Distance1 + Distance2 =125+105=230 miles

7.Sound is said to travel in air at about 1100 feet per second. A man hears the axe striking the tree, 11/5 seconds after he sees it strike the tree. How far is the man from the wood chopper?
A. 1800 ft	B. 2810 ft
C. 3020 ft	D. 2420 ft

 

Answer : Option D

Explanation :

Speed of the sound = 1100 ft/s ⇒Time = ¹¹/₅ second

Distance = Speed × Time = 1100 ×11/5=220×11=2420 ft

8.A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. What is the length of the bridge (in meters)?
A. 1250	B. 1280
C. 1320	D. 1340

 

Answer : Option A

Explanation :

Speed = 5 km/hr

Time = 15 minutes = 1/4 hour

Length of the bridge = Distance Travelled by the man

= Speed × Time = 5×1/4 km

=5×1/4×1000 metre=1250 metre

Level 2

1.A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is
A. 11 hrs	B. 8 hrs 45 min
C. 7 hrs 45 min	D. 9 hts 20 min

 

Answer : Option C

Explanation :

Given that time taken for riding both ways will be 2 hours lesser than the time needed for waking one way and riding back From this, we can understand that time needed for riding one way = time needed for waking one way – 2 hours Given that time taken in walking one way and riding back = 5 hours 45 min Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min In fact, you can do all these calculations mentally and save a lot of time which will be a real benefit for you. 2.A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
A. 121 km	B. 242 km
C. 224 km	D. 112 km

 

Answer : Option C

Explanation :

distance = speed x time
Let time taken to travel the first half = x hr
then time taken to travel the second half = (10 – x) hr
Distance covered in  the first half = 21x
Distance covered in  the second half = 24(10 – x)
But distance covered in  the first half = Distance covered in the second half
=> 21x = 24(10 – x) => 21x = 240 – 24x => 45x = 240 => 9x = 48 => 3x = 16⇒x=16/3

Hence Distance covered in the first half = 21x=21×16/3=7×16=112 km. Total distance = 2×112=224 km

3.A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car?
A. 30 km/hr	B. 35 km/hr
C. 25 km/hr	D. 40 km/hr

 

Answer : Option B

Explanation :

Time = 1 hr 40 min 48 sec = 1hr +40/60hr+48/3600hr=1+2/3+1/75=126/75hr

Distance = 42 kmSpeed=distance/time=42(126/75) =42×75/126

⇒5/7 of the actual speed = 42×75/126

⇒actual speed = 42×75/126×7/5=42×15/18=7×15/3=7×5=35 km/hr

4.A man covered a certain distance at some speed. If he had moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. What is the the distance in km?
A. 36	B. 38
C. 40	D. 42

 

Answer : Option C

Explanation :

Let the distance be x km , the speed in which he moved = v kmph

Time taken when moving at normal speed – time taken when moving 3 kmph faster = 40 minutes

⇒x/v−x/(v+3)=40/60⇒x[1/v−1/(v+3)]=2/3⇒x[(v+3−v)/v(v+3)]=2/3

⇒2v(v+3)=9x…………….(Equation1)

Time taken when moving 2 kmph slower – Time taken when moving at normal speed = 40 minutes
⇒x/(v−2)−x/v=40/60⇒x[1/(v−2)−1/v]=2/3

⇒x[(v−v+2)/v(v−2)]=2/3⇒x[2/v(v−2)]=2/3

⇒x[1/v(v−2)]=1/3⇒v(v−2)=3x…………….(Equation2)

Equation1/Equation2

⇒2(v+3)/(v−2)=3⇒2v+6=3v−6⇒v=12

Substituting this value of v inEquation1⇒2×12×15=9x

=>x= (2×12×15)/9= (2×4×15)/3=2×4×5=40. Hence distance = 40 km

5.In covering a distance of 30 km, Arun takes 2 hours more than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun’s speed?
A. 8 kmph	B. 5 kmph
C. 4 kmph	D. 7 kmph

 

Answer : Option B

Explanation :

Let the speed of Arun = x kmph and the speed of Anil = y kmph
distance = 30 km

We know that distance/speed=time. Hence, 30/x−30/y=2………..(Equation1)

30/y−30/2x=1………..(Equation2)

Equation1 + Equation2⇒30/x−30/2x=3⇒30/2x=3⇒15/x=3⇒5/x=1⇒x=5. Hence Arun’s speed = 5 kmph

6.A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour?
A. 70.24 km/hr	B. 74. 24 km/hr
C. 71.11 km/hr	D. 72.21 km/hr

 

Answer : Option C

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph,the average speed of the whole journey = 2xy/(x+y) kmph.

By using the same formula, we can find out the average speed quickly average speed = (2×64×80)/(64+80)=(2×64×80)/144⇒ (2×32×40)/36= (2×32×10)/9⇒ (64×10)/9=71.11 kmph

7.A man rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. What is his average speed for the entire trip approximately?
A. 11.2 kmph	B. 10 kmph
C. 10.2 kmph	D. 10.8 kmph

 

Answer : Option D

Explanation :

Total distance travelled = 10 + 12 = 22 km

Time taken to travel 10 km at an average speed of 12 km/hr = distance/speed=10/12 hr

Time taken to travel 12 km at an average speed of 10 km/hr = distance/speed=12/10 hr

Total time taken =10/12+12/10 hr

Average speed = distance/time=22/(10/12+12/10)=(22×120)/{(10×10)+(12×12)}

(22×120)/244=(11×120)/122=(11×60)/61=660/61≈10.8 kmph

8.An airplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 123 hours, it must travel at a speed of:
A. 660 km/hr	B. 680 km/hr
C. 700 km/hr	D. 720 km/hr

 

Answer : Option D

Explanation :

Speed and time are inversely proportional ⇒Speed ∝ 1/Time (when distance is constant)

Here distance is constant and Speed and time are inversely proportional

Speed ∝ 1/Time⇒Speed1/Speed2=Time2/Time1

⇒240/Speed2=(1*2/3)5⇒240/Speed2=(5/3)/5⇒240/Speed2=1/3⇒Speed2=240×3=720 km/hr

9.A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. What is the speed of the car?
A. 80 kmph	B. 102 kmph
C. 120 kmph	D. 140 kmph

 

Answer : Option C

Explanation :

Let speed of the car = x kmph

Then speed of the train = x *(100+50)/100=150 x /100=3 x /2 kmph

Time taken by the car to travel from A to B=75/x hours

Time taken by the train to travel from A to B=75/(3 x /2)+12.5/60 hours

Since both start from A at the same time and reach point B at the same time

75/x=75/(3 x /2)+12.5/60⇒25/x=12.5/60⇒x=(25×60)/12.5=2×60=120

TIME AND WORK

In these problems the number of persons, quantity of work done and time taken are important factors. Also time taken by a person depends on the efficiency of that person which comes into picture when different people do the work such as women, children do the work alongside the men. The problems related to time and work can be solved by two major approaches – ratio & proportions and unitary method. Let us proceed to find some formulae related to these questions.

Important Formulas – Time and Work

• If A can do a piece of work in n days, work done by A in 1 day = 1/n

 

• If A does 1/n work in a day, A can finish the work in n days

 

• If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then

M1 D1 H1 / W1 = M2 D2 H2 / W2

 

• If A can do a piece of work in p days and B can do the same in q days, A and B together can finish it in pq / (p+q) days

 

• If A is thrice as good as B in work, then

Ratio of work done by A and B = 3:1

Ratio of time taken to finish a work by A and B = 1: 3

 

SOLVED EXAMPLES

Level 1

1.P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
A. 8/15	B. 7/15
C. 11/15	D. 2/11

 

Answer : Option A

Explanation :

Amount of work P can do in 1 day = 1/15

Amount of work Q can do in 1 day = 1/20

Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60

Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15

Fraction of work left = 1 – 7/15= 8/15

2.A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in — hours.
A. 12 hours	B. 6 hours
C. 8 hours	D. 10 hours

 

Answer : Option A

Explanation :

Work done by A in 1 hour = 1/4

Work done by B and C in 1 hour = 1/3

Work done by A and C in 1 hour = 1/2

Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12

Work done by B in 1 hour = 7/12 – 1/2 = 1/12

=> B alone can complete the work in 12 hours

3.A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
A. 37 ½ days	B. 22 days
C. 31 days	D. 22 days

 

Answer : Option A

Explanation :

Work done by A in 20 days = 80/100 = 8/10 = 4/5

Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 — (1)

Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)

Work done by A and B in 1 day = 1/15 —(2)

Work done by B in 1 day = 1/15 – 1/25 = 2/75

=> B can complete the work in 75/2 days = 37 ½ days

4.P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. How many days does P alone need to finish the remaining work?
A. 8	B. 5
C. 4	D. 6

 

Answer : Option D

Explanation :

Work done by P in 1 day = 1/18

Work done by Q in 1 day = 1/15

Work done by Q in 10 days = 10/15 = 2/3

Remaining work = 1 – 2/3 = 1/3

Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6

5.Anil and Suresh are working on a special assignment. Anil needs 6 hours to type 32 pages on a computer and Suresh needs 5 hours to type 40 pages. If both of them work together on two different computers, how much time is needed to type an assignment of 110 pages?
A. 7 hour 15 minutes	B. 7 hour 30 minutes
C. 8 hour 15 minutes	D. 8 hour 30 minutes

 

Answer : Option C

Explanation :

Pages typed by Anil in 1 hour = 32/6 = 16/3

Pages typed by Suresh in 1 hour = 40/5 = 8

Pages typed by Anil and Suresh in 1 hour = 16/3 + 8 = 40/3

Time taken to type 110 pages when Anil and Suresh work together = 110 × 3 /40 = 33/4

= 8 ¼ hours = 8 hour 15 minutes

6.P works twice as fast as Q. If Q alone can complete a work in 12 days, P and Q can finish the work in — days
A. 1	B. 2
C. 3	D. 4

 

Answer : Option D

Explanation :

Work done by Q in 1 day = 1/12

Work done by P in 1 day = 2 × (1/12) = 1/6

Work done by P and Q in 1 day = 1/12 + 1/6 = ¼

=> P and Q can finish the work in 4 days

7.A work can be finished in 16 days by twenty women. The same work can be finished in fifteen days by sixteen men. The ratio between the capacity of a man and a woman is
A. 1:3	B. 4:3
C. 2:3	D. 2:1

 

Answer : Option B

Explanation :

Work done by 20 women in 1 day = 1/16 Work done by 1 woman in 1 day = 1/(16×20) Work done by 16 men in 1 day = 1/15 Work done by 1 man in 1 day = 1/(15×16) 8.P,Q and R together earn Rs.1620 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day?
A. Rs.40	B. Rs.70
C. Rs.90	D. Rs.100

 

Answer : Option B

Explanation :

Amount Earned by P,Q and R in 1 day = 1620/9 = 180 —(1)

Amount Earned by P and R in 1 day = 600/5 = 120 —(2)

Amount Earned by Q and R in 1 day = 910/7 = 130 —(3)

(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day

– Amount Earned by P,Q and R in 1 day = 120+130-180 = 70

=>Amount Earned by R in 1 day = 70
Ratio of the capacity of a man and woman =1/(15×16) : 1/(16×20) = 1/15 : 1/20

= 1/3 :1/4 = 4:3

Level 2

1.P, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it need to complete the work if P does the work and he is assisted by Q and R on every third day?
A. 10 days	B. 14 days
C. 15 days	D. 9 days

 

Answer : Option C

Explanation :

Amount of work P can do in 1 day = 1/20

Amount of work Q can do in 1 day = 1/30

Amount of work R can do in 1 day = 1/60

P is working alone and every third day Q and R is helping him

Work completed in every three days = 2 × (1/20) + (1/20 + 1/30 + 1/60) = 1/5

So work completed in 15 days = 5 × 1/5 = 1

Ie, the work will be done in 15 days

2.A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in – days if they work together.
A. 18 days	B. 22 ½ days
C. 24 days	D. 26 days

 

Answer : Option B

Explanation :

If A completes a work in 1 day, B completes the same work in 3 days

Hence, if the difference is 2 days, B can complete the work in 3 days

=> if the difference is 60 days, B can complete the work in 90 days

=> Amount of work B can do in 1 day= 1/90

Amount of work A can do in 1 day = 3 × (1/90) = 1/30

Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45

=> A and B together can do the work in 45/2 days = 22 ½ days

3.P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in A. 30 days B. 25 days C. 20 days D. 15 days   Answer : Option B Explanation : Work done by P and Q in 1 day = 1/10 Work done by R in 1 day = 1/50 Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50 But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as Work done by P in 1 day × 2 = 6/50 => Work done by P in 1 day = 3/50 => Work done by Q and R in 1 day = 3/50 Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25 So Q alone can do the work in 25 days 4.6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in – days.
A. 4 days	B. 6 days
C. 2 days	D. 8 days

 

Answer : Option A

Explanation :

Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b

Work done by 6 men and 8 women in 1 day = 1/10

=> 6m + 8b = 1/10

=> 60m + 80b = 1 — (1)

Work done by 26 men and 48 women in 1 day = 1/2

=> 26m + 48b = ½

=> 52m + 96b = 1— (2)

Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200

Work done by 15 men and 20 women in 1 day

= 15/100 + 20/200 =1/4

=> Time taken by 15 men and 20 women in doing the work = 4 days

5.Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?
A. 3 pm	B. 2 pm
C. 1:00 pm	D. 11 am

 

Answer : Option C

Explanation :

Work done by P in 1 hour = 1/8

Work done by Q in 1 hour = 1/10

Work done by R in 1 hour = 1/12

Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120

Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60

From 9 am to 11 am, all the machines were operating.

Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60.

6.A can complete a work in 12 days with a working of 8 hours per day. B can complete the same work in 8 days when working 10 hours a day. If A and B work together, working 8 hours a day, the work can be completed in — days.
A. 5 5⁄11	B. 4 5⁄11
C. 6 4⁄11	D. 6 5⁄11

 

Answer : Option A

Explanation :

A can complete the work in 12 days working 8 hours a day

=> Number of hours A can complete the work = 12×8 = 96 hours

=> Work done by A in 1 hour = 1/96

B can complete the work in 8 days working 10 hours a day

=> Number of hours B can complete the work = 8×10 = 80 hours => Work done by B in 1 hour = 1/80

Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480 => A and B can complete the work in 480/11 hours. A and B works 8 hours a day.

Hence total days to complete the work with A and B working together = (480/11)/ (8) = 60/11 days = 5 ⁵⁄₁₁ days

Pending work = 1- 37/60 = 23/60

Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11

which is approximately equal to 2. Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm

7.If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600.
A. 12	B. 14
C. 16	D. 18

 

Answer : Option C

Explanation :

Wages of 1 woman for 1 day = 21600/(40×30)

Wages of 1 man for 1 day = (21600×2)/(40×30)

Wages of 1 man for 25 days = (21600×2×25)/(40×30)

Number of men = 14400/(21600×2×25)/(40×30)=144/(216×50)/40×30)=144/9=16

8.There is a group of persons each of whom can complete a piece of work in 16 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one more person joins them and this process continues till the work is completed. How many days are needed to complete the work?
A. 3 1⁄4 days	B. 4 1⁄3 days
C. 5 1⁄6 days	D. 6 1⁄5 days

 

Answer : Option C

Explanation :

Work completed in 1st day = 1/16

Work completed in 2nd day = (1/16) + (1/16) = 2/16

Work completed in 3rd day = (1/16) + (1/16) + (1/16) = 3/16

An easy way to attack such problems is from the choices. You can see the choices are

very close to each other. So just see one by one.

For instance, The first choice given in 3 ¹⁄₄

The work done in 3 days = 1/16 + 2/16 + 3/16 = (1+2+3)/16 = 6/16

The work done in 4 days = (1+2+3+4)/16 = 10/16

The work done in 5 days = (1+2+3+4+5)/16 = 15/16, almost close, isn’t it?

The work done in 6 days = (1+2+3+4+5+6)/16 > 1

Hence the answer is less than 6, but greater than 5. Hence the answer is 5 ¹⁄₆ days.

(Just for your reference, work done in 5 days = 15/16)

Pending work in 6th day = 1 – 15/16 = 1/16.

In 6th day, 6 people are working and work done = 6/16.

To complete the work 1/16, time required = (1/16) / (6/16) = 1/6 days.

Hence total time required = 5 + 1/6 = 5 ¹⁄₆ days

 

Introduction

Money is not free and it costs to borrow the money. Normally, the borrower has to pay an extra amount in addition to the amount he had borrowed. i.e, to repay the loan, the borrower has to pay the sum borrowed and the interest.

Lender and Borrower

The person giving the money is called the lender and the person taking the money is the borrower.

Principal (sum)

Principal (or the sum) is the money borrowed or lent out for a certain period. It is denoted by P.

Interest

Interest is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed.

Simple Interest (SI)

If the interest on a sum borrowed for certain period is calculated uniformly, it is called simple interest(SI).

Amount (A)

The total of the sum borrowed and the interest is called the amount and is denoted by A

• The statement “rate of interest 10% per annum” means that the interest for one year on a sum of Rs.100 is Rs.10. If not stated explicitly, rate of interest is assumed to be for one year.

 

• Let Principal = P, Rate = R% per annum and Time = T years. Then

  Simple Interest, SI = PRT/100

 

• From the above formula , we can derive the followings

  P=100×SI/RT

  R=100×SI/PT

  T=100×SI/PR

 

Some Formulae

1. If a sum of money becomes n times in T years at simple interest, then the rate of interest per annum can be given be R = 100(n−1)/T %
2. The annual instalment which will discharge a debt of D due in T years at R% simple interest per annum =100D/ (100T+RT(T-1)/2)
3. If an amount P₁is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by
   R=(P1R1+P2R2)/ (P1+P2)
4. If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, … , R_nrespectively and time periods are T₁, T₂, … , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by (1/R1T1):(1/R2T2):⋯(1/RnTn)
5. If a certain sum of money P lent out for a certain time T amounts to P₁at R₁% per annum and to P₂at R₂% per annum, then P = (P2R1−P1R2)/ (R1−R2) and T = (P1−P2) ×100 years / (P2R1−P1R2)

SOLVED EXAMPLES

LEVEL 1

1.       Arun took a loan of Rs. 1400 with simple interest for as many years as the rate of interest. If he paid Rs.686 as interest at the end of the loan period, what was the rate of interest?
A. 8%	B. 6%
C. 4%	D. 7%

Ans. Let rate = R%

Then, Time, T = R years

P = Rs.1400

SI = Rs.686

SI= PRT/100⇒686 = 1400 × R × R/100⇒686=14 Rx R ⇒49=Rx R ⇒R=7

i.e.,Rate of Interest was 7%. (D)

2.       How much time will it take for an amount of Rs. 900 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
A. 2 years	B. 3 years
C. 1 year	D. 4 years

 

 

Ans. P = Rs.900

SI = Rs.81

T = ?

R = 4.5%

T= 100×SI/PR = 100×81/(900×4.5) = 2 years (A)

3.       A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is :
A. Rs. 700	B. Rs. 690
C. Rs. 650	D. Rs. 698

 

 

Ans. Simple Interest (SI) for 1 year = 854-815 = 39

Simple Interest (SI) for 3 years = 39 × 3 = 117

Principal = 815 – 117 = Rs.698 (D)

 

4.       A sum fetched a total simple interest of Rs. 929.20 at the rate of 8 p.a. in 5 years. What is the sum?
A. Rs. 2323	B. Rs. 1223
C. Rs. 2563	D. Rs. 2353

 

Ans. SI = Rs.929.20

P = ?

T = 5 years

R = 8%

P = 100×SI/RT=100×929.20/(8×5) = Rs.2323 (A)

5.       What will be the ratio of simple interest earned by certain amount at the same rate of interest for 5 years and that for 15 years?
A. 3 : 2	B. 1 : 3
C. 2 : 3	D. 3 : 1

 
Solution 1
Let Principal = P

Rate of Interest = R%

Required Ratio = (PR×5/100)/ (PR×15/100) =1:3 (B)
Solution 2

Simple Interest = PRT100

Here Principal(P) and Rate of Interest (R) are constants

Hence, Simple Interest ∝ T

Required Ratio = Simple Interest for 5 years Simple Interest for 15 years=T1T2=515=13=1:3 (B)

6.       A sum of money amounts to Rs.9800 after 5 years and Rs.12005 after 8 years at the same rate of simple interest. The rate of interest per annum is
A. 15%	B. 12%
C. 8%	D. 5%

 

 

Ans. Simple Interest for 3 years = (Rs.12005 – Rs.9800) = Rs.2205

Simple Interest for 5 years = 22053×5=Rs.3675

Principal (P) = (Rs.9800 – Rs.3675) = Rs.6125

R = 100×SI/PT=100×3675/(6125×5) =12% (B)

7.       A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
A. 5%	B. 10%
C. 7%	D. 8%

 

 

Ans. Let the rate of interest per annum be R%

Simple Interest for Rs. 5000 for 2 years at rate R% per annum +Simple Interest for Rs. 3000 for 4 years at rate R% per annum = Rs.2200

⇒5000×R×2/100+3000×R×4/100=2200

⇒100R + 120R=2200⇒220R=2200⇒R=10

i.e, Rate = 10%. (B)

8.       In how many years, Rs. 150 will produce the same interest at 6% as Rs. 800 produce in 2 years at 4½% ?
A. 4 years	B. 6 years
C. 8 years	D. 9 years

 

 

Ans. Let Simple Interest for Rs.150 at 6% for n years = Simple Interest for Rs.800 at 4½ % for 2 years

150×6×n/100=800×4.5×2/100

150×6×n=800×4.5×2

n=8 years (C)

 

LEVEL 2

1.        Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
A. Rs. 6400	B. Rs. 7200
C. Rs. 6500	D. Rs. 7500

 

 

Ans. Let the investment in scheme A be Rs.x

and the investment in scheme B be Rs. (13900 – x)

We know that SI = PRT/100

Simple Interest for Rs.x in 2 years at 14% p.a. = x×14×2100=28x100Simple Interest for Rs.(13900 – x) in 2 years at 11% p.a. = (13900−x)×11×2/100 =22(13900−x)/100

Total interest =Rs.3508

Thus, 28x/100+22(13900−x)/100 = 3508

28x+305800−22x=350800

6x = 45000

x=45000/6=7500

Investment in scheme B = 13900 – 7500 = Rs.6400 (A)

2.       A certain sum in invested for T years. It amounts to Rs. 400 at 10% per annum. But when invested at 4% per annum, it amounts to Rs. 200. Find the time (T).
A. 45 years	B. 60 years
C. 40 years	D. 50 Years

 
Solution 1
Let the principal = Rs.x

and time = y years

Principal,x amounts to Rs.400 at 10% per annum in y years

Simple Interest = (400-x)

Simple Interest = PRT/100

⇒ (400−x) = x×10×y/100

⇒ (400−x) = xy/10— (equation 1)

Principal,x amounts to Rs.200 at 4% per annum in y years

Simple Interest = (200-x)

Simple Interest = PRT/100

⇒ (200−x) = x×4×y/100

⇒ (200−x) = xy/25— (equation 2)

(equation 1)/(equation2)

⇒(400−x) / (200−x) = (xy/10)/(xy/25)

⇒ (400−x)/ (200−x) =25/10

⇒ (400−x)/ (200−x) =52

⇒800−2x = 1000−5x

⇒200=3x

⇒x =200/3 Substituting this value of x in Equation 1, we get,

(400−200/3) = (200y/3)/10

⇒ (400−200/3) = 20y/3

⇒1200−200=20y

⇒1000=20y

y=1000/20=50 years (D)

Solution 2
If a certain sum of money P lent out for a certain time T amounts to P₁ at R₁% per annum and to P₂ at R₂% per annum, then

P = (P2R1−P1R2)/ (R1−R2)

T = (P1−P2)x 100 years/(P2R1−P1R2)

R₁ = 10%, R₂ = 4%

P₁ = 400, P₂ = 200

T = (P1−P2)x 100 / (P2R1−P1R2) = (400−200)x 100 / (200×10−400×4)

=200 x 100/ (2000−1600) =200 ×100/400 = 12×100=50 years (D)

3.       Mr. Mani invested an amount of Rs. 12000 at the simple interest rate of 10% per annum and another amount at the simple interest rate of 20% per annum. The total interest earned at the end of one year on the total amount invested became 14% per annum. Find the total amount invested.
A. Rs. 25000	B. Rs. 15000
C. Rs. 10000	D. Rs. 20000

 

Ans. If an amount P₁ is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by

R= (P1R1+P2R2)/(P1+P2)

P₁ = Rs. 12000, R₁ = 10%

P₂ =? R₂ = 20%

R = 14%

14 = (12000×10+P2×20)/ (12000+P2)

12000×14+14P2 =120000+20P2

6P2=14×12000−120000=48000

⇒P2=8000

Total amount invested = (P₁ + P₂) = (12000 + 8000) = Rs. 20000 (D)

4.       A sum of money is lent at S.I. for 6 years. If the same amount is paid at 4% higher, Arun would have got Rs. 120 more. Find the principal
A. Rs. 200	B. Rs. 600
C. Rs. 400	D. Rs. 500

 

 

Ans. This means, simple interest at 4% for that principal is Rs.120

P=100×SI/ RT=100×120/ (4×6) =100×30/6 = 100×5 = 500 (D)

5.       The simple interest on Rs. 1820 from March 9, 2003 to May 21, 2003 at 7 1⁄2% rate is
A. Rs. 27.30	B. Rs. 22.50
C. Rs. 28.80	D. Rs. 29

 

 

Ans. Time, T = (22 + 30 + 21) days = 73 days = 73/365 year=1/5 year

Rate, R = 7.5%=15/2%

SI = PRT/100 = 1820× (15/2) × (1/5)/100 = 1820 × (3/2)/100 = 910 × 3/100

= 2730/100 = 27.30 (A)

6.       A sum of Rs. 7700 is to be divided among three brothers Vikas, Vijay and Viraj in such a way that simple interest on each part at 5% per annum after 1, 2 and 3 years respectively remains equal. The Share of Vikas is more than that of Viraj by
A. Rs.1200	B. Rs.1400
C. Rs.2200	D. Rs.2800

Ans. If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, … , R_n respectively and time periods are T₁, T₂, … , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by

1/R1T1:1/R2T2:⋯1/RnTn

 

T₁ = 1 , T₂ = 2, T₃ = 3

R₁ = 5 , R₂ = 5, R₃ = 5

Share of Vikas : Share of Vijay : Share of Viraj

= (1/5×1) : (1/5×2) : (1/5×3) = 1/1:1/2:1/3 = 6:3:2

Total amount is Rs. 7700

Share of Vikas = 7700×6/11=700×6 = 4200

Share of Viraj = 7700×2/11=700×2=1400

Share of Vikas is greater than Share of Viraj by (4200 – 1400) = Rs. 2800 (D)

 

7.       David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?
A. Rs.5000	B. Rs.2000
C. Rs.6000	D. Rs.3000

 

 

Ans. Let x, y and x be his investments in A, B and C respectively. Then

Then, Interest on x at 10% for 1 year

+ Interest on y at 12% for 1 year

+ Interest on z at 15% for 1 year

= 3200

x×10×1/100+y×12×1/100+z×15×1/100=3200

⇒10x+12y+15z=320000−−−(1)

Amount invested in Scheme C was 240% of the amount invested in Scheme B

=>z=240y/100 = 60y/25=12y/5−−−(2)

Amount invested in Scheme C was 150% of the amount invested in Scheme A

=>z=150x/100=3x/2

=>x=2z/3=2/3×12y/5=8y/5−−−(3)

From(1),(2) and (3),

10x + 12y + 15z = 320000

10(8y/5)+12y+15(12y/5)=320000

16y+12y+36y=320000

64y=320000

y=320000/64=10000/2=5000

i.e.,Amount invested in Scheme B = Rs.5000 (A)

 

 

Speed has no sense of direction unlike the velocity. Relative speed is the speed of one object as observed from another moving object. Questions on train are the classic examples of relative speed and in all these questions it is assumed that trains move parallel to each other – whether in the same direction or the opposite direction. Thus, we shall see how the relative speed is calculated and using it we come to know the time taken by the trains to cross each other and some other like aspects.

Important Formulas – Problems on Trains

1. x km/hr = (x×5)/18 m/s

 

2. y m/s = (y×18)/5 km/hr

 

3. Speed = distance/time, that is, s = d/t

 

4. velocity = displacement/time, that is, v = d/t

 

5. Time taken by a train x meters long to pass a pole or standing man or a post
   = Time taken by the train to travel x meters.

 

6. Time taken by a train x meters long to pass an object of length y meters

= Time taken by the train to travel (x + y) metres.

 

7. Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v1 > v2,

then their relative speed = (v1 – v2) m/s

 

8. Suppose two trains or two objects are moving in opposite directions at v1 m/s and v2 m/s ,

then their relative speed = (v1+ v2) m/s

 

9. Assume two trains of length x metres and y metres are moving in opposite directions at v1 m/s and v2 m/s, Then

The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds

 

10. Assume two trains of length x metres and y metres are moving in the same direction at at v1 m/s and v2 m/s where v1 > v2, Then

The time taken by the faster train to cross the slower train = (x+y) / (v1-v2) seconds

 

11. Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then,

A’s speed: B’s speed = √q: √p

 

 

Solved Examples

Level 1

1.A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train?
A. 190 metres	B. 160 metres
C. 200 metres Answer : Option C	D. 120 metres

 

Explanation :

Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s

Time taken to cross, t = 18 s

Distance Covered, d = vt = (400/36)× 18 = 200 m

Distance covered is equal to the length of the train = 200 m

2.A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m?
A. 120 sec	B. 99 s
C. 89 s	D. 80 s

 

Answer : Option C

Explanation :

v = 240/24 (where v is the speed of the train) = 10 m/s

t = (240+650)/10 = 89 seconds

3.Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is
A. 10.8 s	B. 12 s
C. 9.8 s	D. 8 s

 

Answer : Option A

Explanation :

Distance = 140+160 = 300 m

Relative speed = 60+40 = 100 km/hr = (100×10)/36 m/s

Time = distance/speed = 300 / (100×10)/36 = 300×36 / 1000 = 3×36/10 = 10.8 s

4.A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?
A. 79.2 km/hr	B. 69 km/hr
C. 74 km/hr	D. 61 km/hr

 

Answer : Option A

Explanation :

Let x is the length of the train and v is the speed

Time taken to move the post = 8 s

=> x/v = 8

=> x = 8v — (1)

Time taken to cross the platform 264 m long = 20 s

(x+264)/v = 20

=> x + 264 = 20v —(2)

Substituting equation 1 in equation 2, we get

8v +264 = 20v

=> v = 264/12 = 22 m/s

= 22×36/10 km/hr = 79.2 km/hr

5.Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is
A. 2 : 3	B. 2 :1
C. 4 : 3	D. 3 : 2

 

Answer : Option C

Explanation :

Ratio of their speeds = Speed of first train : Speed of second train

= √16: √ 9

= 4:3

6.Train having a length of 270 meter is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
A. 320 m	B. 190 m
C. 210 m	D. 230 m

 

Answer : Option D

Explanation :

Relative speed = 120+80 = 200 kmph = 200×10/36 m/s = 500/9 m/s

time = 9s

Total distance covered = 270 + x where x is the length of other train

(270+x)/9 = 500/9

=> 270+x = 500

=> x = 500-270 = 230 meter

7.Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?
A. 10.30 a.m	B. 10 a.m.
C. 9.10 a.m.	D. 11 a.m.

 

Answer : Option B

Explanation :

Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3 Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

8.Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is
A. 42	B. 36
C. 28	D. 20

 

Answer : Option B

Explanation :

Distance covered = 120+120 = 240 m

Time = 12 s

Let the speed of each train = v. Then relative speed = v+v = 2v

2v = distance/time = 240/12 = 20 m/s

Speed of each train = v = 20/2 = 10 m/s

= 10×36/10 km/hr = 36 km/hr

 

Level 2

1.A train, 130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is
A. 270 m	B. 245 m
C. 235 m	D. 220 m

 

Answer : Option B

Explanation :

Assume the length of the bridge = x meter

Total distance covered = 130+x meter

total time taken = 30s

speed = Total distance covered /total time taken = (130+x)/30 m/s

=> 45 × (10/36) = (130+x)/30

=> 45 × 10 × 30 /36 = 130+x

=> 45 × 10 × 10 / 12 = 130+x

=> 15 × 10 × 10 / 4 = 130+x

=> 15 × 25 = 130+x = 375

=> x = 375-130 =245

2.A train has a length of 150 meters. It is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train.
A. 182 km/hr	B. 180 km/hr
C. 152 km/hr	D. 169 km/hr

 

Answer : Option A

Explanation :

Length of the train, l = 150m

Speed of the man, Vm= 2 km/hr

Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr

Relative Speed = Speed of train, Vt – Speed of man (As both are moving in the same direction)

=> 180 = Vt – 2 => Vt = 180 + 2 = 182 km/hr

3.Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively. If they cross each other in 23 seconds, what is the ratio of their speeds?
A. Insufficient data	B. 3 : 1
C. 1 : 3	D. 3 : 2

 

Answer : Option D

Explanation :

Let the speed of the trains be x and y respectively

length of train1 = 27x

length of train2 = 17y

Relative speed= x+ y

Time taken to cross each other = 23 s

=> (27x + 17 y)/(x+y) = 23 => (27x + 17 y)/ = 23(x+y)

=> 4x = 6y => x/y = 6/4 = 3/2

4.A jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. How much time does it take for the train to pass the jogger?
A. 46	B. 36
C. 18	D. 22

 

Answer : Option B

Explanation :

Distance to be covered = 240+ 120 = 360 m

Relative speed = 36 km/hr = 36×10/36 = 10 m/s

Time = distance/speed = 360/10 = 36 seconds

5.A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is
A. None of these	B. 280 meter
C. 240 meter	D. 200 meter

 

Answer : Option C

Explanation :

Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s

Length of the train = speed × time taken to cross the man = 15×20 = 300 m

Let the length of the platform = L

Time taken to cross the platform = (300+L)/15

=> (300+L)/15 = 36

=> 300+L = 15×36 = 540 => L = 540-300 = 240 meter

6.A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train?
A. 62 m	B. 54 m
C. 50 m	D. 55 m

 

Answer : Option C

Explanation :

Let x is the length of the train in meter and v is its speed in kmph

x/9 = (v-2) (10/36) — (1)

x/10 = (v-4) (10/36) — (2)

Dividing equation 1 with equation 2

10/9 = (v-2)/(v-4) => 10v – 40 = 9v – 18 => v = 22

Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m

7.A train is traveling at 48 kmph. It crosses another train having half of its length, traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform?
A. 500 m	B. 360 m
C. 480 m	D. 400 m

 

Answer : Option D

Explanation :

Speed of train1 = 48 kmph

Let the length of train1 = 2x meter

Speed of train2 = 42 kmph

Length of train 2 = x meter (because it is half of train1’s length)

Distance = 2x + x = 3x

Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s

Time = 12 s

Distance/time = speed => 3x/12 = 25

=> x = 25×12/3 = 100 meter

Length of the first train = 2x = 200 meter

Time taken to cross the platform= 45 s

Speed of train1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y where y is the length of the platform

=> 200 + y = 45×40/3 = 600

=> y = 400 meter

8.A train, 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)?
A. 440 m	B. 500 m
C. 260 m	D. 430 m

 

Answer : Option B

Explanation :

Distance = 800+x meter where x is the length of the tunnel

Time = 1 minute = 60 seconds

Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s

Distance/time = speed

(800+x)/60 = 65/3 => 800+x = 20×65 = 1300

=> x = 1300 – 800 = 500 meter

9.Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely passes a man sitting in the slower train in 5 seconds, the length of the fast train is :
A. 19 m	B. 2779 m
C. 1329 m	D. 33 m

 

Answer : Option B

Explanation :

Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s

Time = 5 s

Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = 2779 m = length of the fast train

 

• Percentage

  Percent means for every 100

  So, when percent is calculated for any value, it means that we calculate the value for every 100 of the reference value.

  percent is denoted by the symbol %. For example, x percent is denoted by x%

• x%=x/100

  Example : 25%=25/100=1/4

• To express x/y as a percent,we have x/y=(x/y×100)%

  Example : 1/4=(1/4×100)%=25%

• If the price of a commodity increases by R%, the reduction in consumptionso as not to increase the expenditure = [R/(100+R)×100]%
• If the price of a commodity decreases by R%, the increase in consumptionso as not to decrease the expenditure = [R/(100−R)×100]%
• If the population of a town = P and it increases at the rate of R% per annum, thenPopulation after n years = P((1+R)/100))n
• If the population of a town = P and it increases at the rate of R% per annum, thenPopulation before n years = P((1+R)/100))n
• If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine after n years = P((1-R)/100))n

• If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine before n years = P((1-R)/100))n

 

Solved Examples

Level 1

1.    If A = x% of y and B = y% of x, then which of the following is true?
A. None of these	B. A is smaller than B.
C. Relationship between A and B cannot be determined.	D. If x is smaller than y, then A is greater than B.
E. A is greater than B.

 

Answer : Option A

Explanation :

A = xy/100 ………….(Equation 1)

B = yx/100……………..(Equation 2)

From these equations, it is clear that A = B

 

 

2.If 20% of a = b, then b% of 20 is the same as:
A. None of these	B. 10% of a
C. 4% of a	D. 20% of a

 

Answer :Option C

Explanation :

20% of a = b

=> b = 20a/100

b% of 20 = 20b/100=(20a/100) × 20/100

=(20×20×a)/(100×100)=4a/100 = 4% of a

 

3.Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.
A. 2 : 1	B. 1 : 2
C. 1 : 1	D. 4 : 3

 

 

Answer :Option D

Explanation :

5% of A + 4% of B = 2/3(6% of A + 8% of B)

5A/100+4B/100=2/3(6A/100+8B/100)

⇒5A+4B=2/3(6A+8B)

⇒15A+12B=12A+16B

⇒3A=4B

⇒AB=43⇒A:B=4:3

4.The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year?
A. 4%	B. 6%
C. 5%	D. 50%

 

Answer :Option C

Explanation :

Increase in the population in 10 years = 2,62,500 – 1,75,000 = 87500

% increase in the population in 10 years = (87500/175000)×100=8750/175=50%

Average % increase of population per year = 50%/10=5%

 

5.Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
A. 57%	B. 50%
C. 52%	D. 60%

 

Answer :Option A

Explanation :

Votes received by the winning candidate = 11628

Total votes = 1136 + 7636 + 11628 = 20400

Required percentage = (11628/20400)×100=11628/204=2907/51=969/17=57%

 

6.A fruit seller had some oranges. He sells 40% oranges and still has 420 oranges. How many oranges he had originally?
A. 420	B. 700
C. 220	D. 400

 

Answer :Option B

Explanation :

He sells 40% of oranges and still there are 420 oranges remaining

=> 60% of oranges = 420

⇒(60×Total Oranges)/100=420

⇒Total Oranges/100=7

⇒ Total Oranges = 7×100=700

7.A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
A. 499/11 %	B. 45 %
C. 500/11 %	D. 489/11 %

 

Answer :Option C

Explanation :

Total runs scored = 110

Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60

Total runs scored by running between the wickets = 110 – 60 = 50

Required % = (50/110)×100=500/11%

 

8.What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
A. 2023%	B. 20%
C. 21%	D. 2223%

 

 

Answer :Option B

Explanation :

Total numbers = 70

Total numbers in 1 to 70 which has 1 in the unit digit = 7

Total numbers in 1 to 70 which has 9 in the unit digit = 7

Total numbers in 1 to 70 which has 1 or 9 in the unit digit = 7 + 7 = 14

Required percentage = (14/70)×100=140/7=20%

 

Level 2

 

1.In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got?
A. 2800	B. 2700
C. 2100	D. 2500

 

Answer :Option B

Explanation :

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = (7500×80)/100

1^st candidate got 55% of the total valid votes.

Hence the 2^nd candidate should have got 45% of the total valid votes
=> Valid votes that 2nd candidate got = (total valid votes ×45)/100

=7500×(80/100)×(45/100)=75×(4/5)×45=75×4×9=300×9=2700

 

2.In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State?
A. 8200	B. 7500
C. 7000	D. 8000

 

Answer :Option D

Explanation :

State A and State B had an equal number of candidates appeared.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

 

3.In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school?
A. 100	B. 102
C. 110	D. 90

 

 

Answer :Option A

Explanation :

Let the total number of students = x

Given that 20% of students are below 8 years of age

then The number of students above or equal to 8 years of age = 80% of x —–(Equation 1)

Given that number of students of 8 years of age = 48 —–(Equation 2)

Given that number of students above 8 years of age = 2/3 of number of students of 8 years of age

=>number of students above 8 years of age = (2/3)×48=32—–(Equation 3)

From Equation 1,Equation 2 and Equation 3,
80% of x = 48 + 32 = 80

⇒80x/100=80

⇒x100=1⇒x=100

4.In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examination?
A. 28000	B. 30000
C. 32000	D. 33000

 

Answer :Option B

Explanation :

Let the number of candidates applied for the examination = x

Given that 5% of the applicants were found ineligible.

It means that 95% of the applicants were eligible (∴ 100% – 5% = 95%)

Hence total eligible candidates = 95x/100

Given that 85% of the eligible candidates belonged to the general category

It means 15% of the eligible candidates belonged to other categories(∴ 100% – 85% = 15%)
Hence Total eligible candidates belonged to other categories

=(total eligible candidates×15)/100=(95x/100)×(15/100)

=(95x×15)/(100×100)

Given that Total eligible candidates belonged to other categories = 4275

⇒(95x×15)/(100×100)=4275

⇒(19x×15)/(100×100)=855

⇒(19x×3)/(100×100)=171

⇒(x×3)/(100×100)=9

⇒x/(100×100)=3

⇒x=3×100×100=30000

 

5.A student multiplied a number by 3/5 instead of 5/3.What is the percentage error in the calculation?
A. 64%	B. 32%
C. 34%	D. 42%

 

Answer :Option A

Explanation :

Let the number = 1

Then, ideally he should have multiplied 1 by 5/3.

Hence the correct result was 1 x (5/3) = (5/3)

By mistake, he multiplied 1 by 3/5.

Hence the result with the error = 1 x (3/5) = (3/5)

Error = 5/3−3/5=(25−9)/15=16/15

percentage error = (Error/True Value)×100={(16/15)/(5/3)}×100

=(16×3×100)/(15×5)=(16×100)/(5×5)=16×4=64%

 

6.The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ?
A. Rs. 76,375	B. Rs. 34,000
C. Rs. 82,150	D. Rs. 70,000

 

Answer :Option A

Explanation :

Price of the car = Rs.3,25,000

Car insured to 85% of its price

=>Insured price=(325000×85)/100

Insurance company paid 90% of the insurance
⇒Amount paid by Insurance company =(Insured price×90)/100

=325000×(85/100)×(90/100)=325×85×9=Rs.248625

Difference between the price of the car and the amount received

= Rs.325000 – Rs.248625 = Rs.76375

7.If the price of petrol increases by 25% and Benson intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased?
A. 8%	B. 7%
C. 10%	D. 6%

 

 

Answer :Option A

Explanation :

Assume that the initial price of 1 Litre petrol = Rs.100 ,Benson spends Rs.100 for petrol,

such that Benson buys 1 litre of petrol

After the increase by 25%, price of 1 Litre petrol = (100×(100+25))/100=Rs.125

Since Benson spends additional 15% on petrol,

amount spent by Benson = (100×(100+15))/100=Rs.115

Hence Quantity of petrol that he can purchase = 115/125 Litre

Quantity of petrol reduced = (1−115/125) Litre

Percentage Quantity of reduction = ((1−115/125))/1×100=(10/125)/×100=(10/5)×4=2×4=8%

8.30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
A. 60%	B. 70%
C. 80%	D. 90%

 

Answer :Option C

Explanation :

Let total number of men = 100

Then

80 men are less than or equal to 50 years old

(Since 80% of the men are less than or equal to 50 years old)

=> 20 men are above 50 years old (Since we assumed total number of men as 100)

20% of the men above the age of 50 play football

⇒Number of men above the age of 50 who play football = (20×20)/100=4

Number of men who play football = 20 (Since 20% of all men play football)
Percentage of men who play football above the age of 50 = (4/20)×100=20%

=>Percentage of men who play football less than or equal to the age 50 = 100%−20%=80%

 

1. Alligation

   It is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.

2. Mean Price

   Mean price is the cost price of a unit quantity of the mixture

 

3. Suppose a container contains x of liquid from which y units are taken out and replaced by water.After n operations, the quantity of pure liquid = [x(1−y/x)n]

 

4. Rule of Alligation

   If two ingredients are mixed, then

   (Quantity of cheaper/Quantity of dearer)=(P. of dearer – Mean Price)/(Mean price – C.P. of cheaper)

Cost Price(CP) of a unit quantity of cheaper (c)	Cost Price(CP) of a unit quantity of dearer (d)
Mean Price
(m)
(d – m)	(m – c)

5. => (Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)

 

Solved Examples

Level 1

1.A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A. 26 litres	B. 29.16 litres
C. 28 litres	D. 28.2 litres

 

 

Answer : Option B

Explanation :

Assume that a container contains x of liquid from which y units are taken out and replaced

by water. After n operations, the quantity of pure liquid

=[x(1−y/x)n]

Hence milk now contained by the container = 40(1−4/40)3=40(1−1/10)3

40×9/10×9/10×9/10=(4×9×9×9)/100=29.16

2.A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is
A. 43	B. 34
C. 32	D. 23

 

 

Answer : Option D

Explanation :

Concentration of alcohol in 1st Jar = 40%

Concentration of alcohol in 2nd Jar = 19%

After the mixing, Concentration of alcohol in the mixture = 26%

By the rule of alligation,

Concentration of alcohol in 1st Jar	Concentration of alcohol in 2nd Jar
40%	19%
Mean
26%
7	14

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

 

3.In what ratio should rice at Rs.9.30 per Kg be mixed with rice at Rs. 10.80 per Kg so that the mixture be worth Rs.10 per Kg ?
A. 7 : 8	B. 8 : 7
C. 6 : 7	D. 7 ; 6

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

 

Cost of 1 kg rice of 1st kind	Cost of 1 kg rice of 2nd kind
9.3	10.80
Mean Price
10
10.8-10 = .8	10 – 9.3 = .7

Required ratio = .8 : .7 = 8 : 7.

 

 

 

4.In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre?
A. 1 : 3	B. 2 : 2
C. 1 : 2	D. 3 : 1

 

 

Answer : Option C

 

Explanation :

By the rule of alligation, we have

Cost Price of 1 litre of water	Cost Price of 1 litre of milk
0	12
Mean Price
8
12-8=4	8-0=8

Required Ratio = 4 : 8 = 1 : 2

 

 

 

5.In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10%
A. 1.5 Kg	B. 2 Kg
C. .5 Kg	D. 1 Kg

 

 

 

 

Answer : Option D

 

Explanation :

By the rule of alligation, we have

Percentage concentration of manganese in the mixture : 20	Percentage concentration of manganese in pure iron : 0
Percentage concentration of manganese in the final mixture
10
10 – 0 = 10	20 – 10 = 10

=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1

Given that Quantity of the mixture = 1 Kg

Hence Quantity of iron to be added = 1 Kg

 

6.A trader has 1600Kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%.
A. 1200 Kg	B. 1400 Kg
C. 1600 Kg	D. 800 Kg

 

 

 

Answer : Option A

 

Explanation :

By the rule of alligation, we have

 

% Profit by selling part1	% Profit by selling part2
8	12
Net % Profit
11
12 – 11 = 1	11 – 8 = 3

=>Quantity of part1 : Quantity of part2 = 1 : 3

Given that total quantity = 1600 Kg

Hence quantity of part2 (quantity sold at 12%) = 1600×3/4=1200

 

 

7.How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it
A. 4 litre	B. 2 litre
C. 1 litre	D. 3 litre

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

% Concentration of water in pure water : 100	% Concentration of water in the given mixture : 10
Mean % concentration
20
20 – 10 = 10	100 – 20 = 80

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here Quantity of the mixture = 16 litres

=> Quantity of water : 16 = 1 : 8

Quantity of water = 16×1/8=2 litre

 

8.A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is
A. 300	B. 400
C. 600	D. 500

 

 

Answer : Option C

 

Explanation :

By the rule of alligation,we have

Profit% by selling 1st part	Profit% by selling 2nd part
8	18
Net % profit
14
18-14=4	14-8=6

=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3

Total quantity is given as 1000Kg

So Quantity of part2 (Quantity sold at 18% profit) = 1000×3/5=600Kg

 

Level 2

 

1.Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ?
A. Rs.182.50	B. Rs.170.5
C. Rs.175.50	D. Rs.180

 

 

Answer : Option C

Explanation :

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1

So their average price = (126+135)2=130.5

Hence let’s consider that the mixture is formed by mixing two varieties of tea.

one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e.,

1 : 1. Now let’s find out x.

By the rule of alligation, we can write as

Cost of 1 kg of 1st kind of tea	Cost of 1 kg of 2nd kind of tea
130.50	x
Mean Price
153
(x – 153)	22.50

(x – 153) : 22.5 = 1 : 1

=>x – 153 = 22.50

=> x = 153 + 22.5 = 175.5

 

2.A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 5litres, 7 litres	B. 7litres, 4 litres
C. 6litres, 6 litres	D. 4litres, 8 litres

 

 

Answer : Option C

Explanation :

Let cost of 1 litre milk be Rs.1

Milk in 1 litre mix in 1st can = 3/4 litre

Cost Price(CP) of 1 litre mix in 1st can = Rs.3/4

Milk in 1 litre mix in 2nd can = 1/2 litre

Cost Price(CP) of 1 litre mix in 2nd can = Rs.1/2

Milk in 1 litre of the final mix=5/8

Cost Price(CP) of 1 litre final mix = Rs.5/8

=> Mean price = 5/8
By the rule of alligation, we can write as

CP of 1 litre mix in 2nd can	CP of 1 litre mix in 1st can
1/2	3/4
Mean Price
5/8
3/4 – 5/8 = 1/8	5/8 – 1/2 = 1/8

=> mix in 2nd can :mix in 1st can = 1/8 : 1/8 = 1:1

ie, from each can, 12×12=6 litre should be taken

 

3.Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ?
A. 3: 4	B. 4 : 3
C. 9 : 7	D. 7 : 9

 

 

Answer : Option D

 

Explanation :

 

Let Cost Price(CP) of 1 litre spirit be Rs.1

Quantity of spirit in 1 litre mixture from vessel A = 5/7

Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7

Quantity of spirit in 1 litre mixture from vessel B = 7/13

Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13

Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13

Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price

By the rule of alligation, we can write as

 

CP of 1 litre mixture from vessel A	CP of 1 litre mixture from vessel B
5/7	7/13
Mean Price
8/13
8/13 – 7/13 = 1/13	5/7 – 8/13 = 9/91

=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio

 

4.How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?
A. 60 Kg	B. 63 kg
C. 58 Kg	D. 56 Kg

 

 

Answer : Option B

Explanation :

Selling Price(SP) of 1 Kg mixture= Rs. 9.24

Profit = 10%

Cost Price(CP) of 1 Kg mixture= 100×SP /(100+Profit%) =100×9.24/(100+10)

=100×9.24/110=92.4/11=Rs.8.4

By the rule of alligation, we have

CP of 1 kg sugar of 1st kind	CP of 1 kg sugar of 2nd kind
Rs. 9	Rs. 7
Mean Price
Rs.8.4
8.4 – 7 = 1.4	9 – 8.4 = .6

ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the

ratio 1.4 : .6 = 14 : 6 = 7 : 3

Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar

then x : 27 = 7 : 3

⇒x/27=7/3

⇒x=(27×7)/3=63

5.A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially?
A. 23	B. 21
C. 19	D. 17

 

 

Answer : Option B

Explanation :

Let’s initial quantity of P in the container be 7x

and initial quantity of Q in the container be 5x

Now 9 litres of mixture are drawn off from the container

Quantity of P in 9 litres of the mixtures drawn off = 9×7/12=63/12=21/ 4

Quantity of Q in 9 litres of the mixtures drawn off = 9×5/12=45/12=1/54

HenceQuantity of P remains in the mixtures after 9 litres is drawn off =7x−21/4

Quantity of Q remains in the mixtures after 9 litres is drawn off =5x−15/4

Since the container is filled with Q after 9 litres of mixture is drawn off,

Quantity of Q in the mixtures=5x−15/4+9=5x+21/4

Given that the ratio of P and Q becomes 7 : 9

⇒(7x−21/4):(5x+21/4)=7:9

⇒(7x−21/4)(5x+21/4)=7/9

63x−(9×21/4)=35x+(7×2/14)

28x=(16×21/4)

x=(16×21)/(4×28)

litres of P contained in the container initially = 7x=(7×16×21)/(4×28)=(16×21)/(4×4)=21

 

6.A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
A. 25%	B. 20%
C. 22%	D. 24%

 

 

Answer : Option B

Explanation :

Let CP of 1 litre milk = Rs.1

Given that SP of 1 litre mixture = CP of 1 Litre milk = Rs.1

Given than Gain = 25%

Hence CP of 1 litre mixture = 100/(100+Gain%)×SP

=100(100+25)×1

=100/125=4/5

By the rule of alligation, we have

CP of 1 litre milk	CP of 1 litre water
1	0
CP of 1 litre mixture
4/5
4/5 – 0 = 4/5	1- 4/5 = 1/5

=> Quantity of milk : Quantity of water = 4/5 : 1/5 = 4 : 1

Hence percentage of water in the mixture = 1/5×100=20%

 

Mensuration is the branch of mathematics which deals with the study of different geometrical shapes, their areas and Volume. In the broadest sense, it is all about the process of measurement. It is based on the use of algebraic equations and geometric calculations to provide measurement data regarding the width, depth and volume of a given object or group of objects

• Pythagorean Theorem (Pythagoras’ theorem)

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides

c² = a² + b² where c is the length of the hypotenuse and a and b are the lengths of the other two sides

• Pi is a mathematical constant which is the ratio of a circle’s circumference to its diameter. It is denoted by π

π≈3.14≈227

• Geometric Shapes and solids and Important Formulas

Geometric Shapes	Description	Formulas
	Rectangle l = Length b = Breadth d= Length of diagonal	Area = lb Perimeter = 2(l + b) d = √l2+b2
	Square a = Length of a side d= Length of diagonal	Area= a*a=1/2*d*d Perimeter = 4a d = 2√a
	Parallelogram b and c are sides b = base h = height	Area = bh Perimeter = 2(b + c)
	Rhombus a = length of each side b = base h = height d1, d2 are the diagonal	Area = bh(Formula 1) Area = ½*d1*d2 (Formula 2 ) Perimeter = 4a
	Triangle a , b and c are sides b = base h = height	Area = ½*b*h (Formula 1) Area(Formula 2)                         = √S(S−a)(S−b)(S−c              where S is the semiperimeter S  =(a+b+c)/2 (Formula 2 for area          -Heron’s formula) Perimeter = a + b + c Radius of incircle of a triangle of area A =AS where S is the semiperimeter =(a+b+c)/2
	Equilateral Triangle a = side	Area = (√3/4)*a*a               Perimeter = 3a Radius of incircle of an equilateral                                                                  triangle of side a = a/2*√3 Radius of circumcircle of an equilateral triangle of side a = a/√3
Base a is parallel to base b	Trapezium(Trapezoid in American English) h = height	Area = 12(a+b)h
	Circle r = radius d = diameter	d = 2r Area = πr2 = 14πd2 Circumference = 2πr = πd
	Sector of Circle r = radius θ = central angle	Area  = (θ/360) *π*r*r Arc Length, s = (θ/180)* π*r In the radian system for angular measurement, 2π radians = 360° => 1 radian = 180°π => 1° = π180 radians Hence, Angle in Degrees = Angle in Radians × 180°π Angle in Radians = Angle in Degrees × π180°
	Ellipse Major axis length = 2a Minor axis length = 2b	Area = πab Perimeter ≈
	Rectangular Solid l = length w = width h = height	Total Surface Area = 2lw + 2wh + 2hl = 2(lw + wh + hl) Volume = lwh
	Cube s = edge	Total Surface Area = 6s2 Volume = s3
	Right Circular Cylinder h = height r = radius of base	Lateral Surface Area = (2 π r)h Total Surface Area = (2 π r)h + 2 (π r2) Volume = (π r2)h
	Pyramid h = height B = area of the base	Total Surface Area = B +                 Sum of  the areas of the triangular sides Volume = 1/3*B*h
	Right Circular Cone h = height r = radius of base	Lateral Surface Area=πrs where s is the slant height =√r*r+h*h Total Surface Area                                 =πrs+πr2
	Sphere r = radius d = diameter	d = 2r Surface Area =4πr*r=πd*d Volume =4/3πr*r*r=16πd*d*d

 

• Important properties of Geometric Shapes
  1. Properties of Triangle
     1. Sum of the angles of a triangle = 180°
     2. Sum of any two sides of a triangle is greater than the third side.

• The line joining the midpoint of a side of a triangle to the positive vertex is called the median

1. The median of a triangle divides the triangle into two triangles with equal areas
2. Centroid is the point where the three medians of a triangle meet.
3. Centroid divides each median into segments with a 2:1 ratio

• Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle.
• An equilateral triangle is a triangle in which all three sides are equal

1. In an equilateral triangle, all three internal angles are congruent to each other
2. In an equilateral triangle, all three internal angles are each 60°
3. An isosceles triangle is a triangle with (at least) two equal sides

• In isosceles triangle, altitude from vertex bisects the base.

 

1. Properties of Quadrilaterals
2. Rectangle
   1. The diagonals of a rectangle are equal and bisect each other
   2. opposite sides of a rectangle are parallel

• opposite sides of a rectangle are congruent

1. opposite angles of a rectangle are congruent
2. All four angles of a rectangle are right angles
3. The diagonals of a rectangle are congruent
4. Square

• All four sides of a square are congruent
• Opposite sides of a square are parallel

1. The diagonals of a square are equal
2. The diagonals of a square bisect each other at right angles
3. All angles of a square are 90 degrees.

• A square is a special kind of rectangle where all the sides have equal length

1. Parallelogram

• The opposite sides of a parallelogram are equal in length.
• The opposite angles of a parallelogram are congruent (equal measure).

1. The diagonals of a parallelogram bisect each other.

• Each diagonal of a parallelogram divides it into two triangles of the same area

1. Rhombus

• All the sides of a rhombus are congruent
• Opposite sides of a rhombus are parallel.
• The diagonals of a rhombus bisect each other at right angles

1. Opposite internal angles of a rhombus are congruent (equal in size)

• Any two consecutive internal angles of a rhombus are supplementary; i.e. the sum of their angles = 180° (equal in size)
• If each angle of a rhombus is 90°, it is a square

Other properties of quadrilaterals

• The sum of the interior angles of a quadrilateral is 360 degrees
• If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)
• A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
• Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.
• Each diagonal of a parallelogram divides it into two triangles of the same area
• A square is a rhombus and a rectangle.

1. Sum of Interior Angles of a polygon
   3. The sum of the interior angles of a polygon = 180(n – 2) degrees where n = number of sides Example 1 : Number of sides of a triangle = 3. Hence, sum of the interior angles of a triangle = 180(3 – 2) = 180 × 1 = 180 ° Example 2 : Number of sides of a quadrilateral = 4. Hence, sum of the interior angles of any quadrilateral = 180(4 – 2) = 180 × 2 = 360.

 

 

Solved Examples

Level 1

1. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?
2. 4.04 %
3. 2.02 %
4. 4 %
5. 2 %

Answer : Option A

Explanation :

Error = 2% while measuring the side of a square.

Let the correct value of the side of the square = 100
Then the measured value = (100×(100+2))/100=102 (∵ error 2% in excess)

Correct Value of the area of the square = 100 × 100 = 10000
Calculated Value of the area of the square = 102 × 102 = 10404

Error = 10404 – 10000 = 404
Percentage Error = (Error/Actual Value)×100=(404/10000)×100=4.04%

 

2. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area?
3. 30 %
4. 28 %
5. 32 %
6. 26 %

Answer : Option B

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Lost 20% of length
=> New length =( Original length × (100−20))/100
=(100×80)/100=80

Lost 10% of breadth
=> New breadth= (Original breadth × (100−10))/100
=(100×90)/100=90

New area = 80 × 90 = 7200

Decrease in area
= Original Area – New Area
= 10000 – 7200 = 2800

Percentage of decrease in area
=(Decrease in Area/Original Area)×100=(2800/10000)×100=28%

4. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?
5. 25 % Increase
6. 25 % Decrease
7. 50 % Decrease
8. 50 % Increase

Answer : Option D

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Length of the rectangle is halved
=> New length = (Original length)/2=100/2=50

breadth is tripled
=> New breadth= Original breadth × 3 = 100 × 3 = 300

New area = 50 × 300 = 15000

Increase in area = New Area – Original Area = 15000 – 10000= 5000
Percentage of Increase in area =( Increase in Area/OriginalArea)×100=(5000/10000)×100=50%

5. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot?
6. 14 metres
7. 20 metres
8. 18 metres
9. 12 metres

Answer : Option B

Explanation:

lb = 460 m² ——(Equation 1)

Let the breadth = b
Then length, l =( b×(100+15))/100=115b/100——(Equation 2)

From Equation 1 and Equation 2,
115b/100×b=460b2=46000/115=400⇒b=√400=20 m

 

6. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus?
7. equal to ½
8. equal to ¾
9. greater than 1
10. equal to 1

Answer : Option C

Explanation :

If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)

 

Hence greater than 1 is the more suitable choice from the given list

================================================================
Note : Proof

Consider a square and rhombus standing on the same base ‘a’. All the sides of a square are of equal length. Similarly all the sides of a rhombus are also of equal length. Since both the square and rhombus stands on the same base ‘a’,

Length of each side of the square = a
Length of each side of the rhombus = a

Area of the sqaure = a² …(1)

From the diagram, sin θ = h/a
=> h = a sin θ

Area of the rhombus = ah = a × a sin θ = a² sin θ …(2)

From (1) and (2)

Area of the square/Area of the rhombus= a² /a²sinθ=1/sinθ

Since 0° < θ < 90°, 0 < sin θ < 1. Therefore, area of the square is greater than that of rhombus, provided both stands on same base.

(Note that, when each angle of the rhombus is 90°, rhombus is also a square (can be considered as special case) and in that case, areas will be equal.

 

7. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field.
8. 37500 m²
9. 30500 m²
10. 32500 m²
11. 40000 m²

Answer : Option A

Explanation :

Given that breadth of a rectangular field is 60% of its length
⇒b=(60/100)* l =(3/5)* l

perimeter of the field = 800 m
=> 2 (l + b) = 800
⇒2(l+(3/5)* l)=800⇒l+(3/5)* l =400⇒(8/5)* l =400⇒l/5=50⇒l=5×50=250 m

b = (3/5)* l =(3×250)/5=3×50=150 m

Area = lb = 250×150=37500 m2

 

8. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%?
9. 45%
10. 44%
11. 40%
12. 42%

Answer : Option B

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Increase in 20% of length.
=> New length = (Original length ×(100+20))/100=(100×120)/100=120

Increase in 20% of breadth
=> New breadth= (Original breadth × (100+20))/100=(100×120)/100=120

New area = 120 × 120 = 14400

Increase in area = New Area – Original Area = 14400 – 10000 = 4400
Percentage increase in area =( Increase in Area /OriginalArea)×100=(4400/10000)×100=44%

9. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
10. 814
11. 802
12. 836
13. 900

Answer : Option A

Explanation :

l = 15 m 17 cm = 1517 cm
b = 9 m 2 cm = 902 cm
Area = 1517 × 902 cm²

Now we need to find out HCF(Highest Common Factor) of 1517 and 902.
Let’s find out the HCF using long division method for quicker results

902)  1517  (1

 

-902

—————–

615)  902  (1

 

• 615

————–

 

287)  615 (2

 

-574

—————–

 

41)  287  (7

 

-287

————

0
————

Hence, HCF of 1517 and 902 = 41

Hence, side length of largest square tile we can take = 41 cm
Area of each square tile = 41 × 41 cm²

Number of tiles required = (1517×902)/(41×41)=37×22=407×2=814

 

Level 2

1. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet?
2. 126 sq. ft.
3. 64 sq. ft.
4. 100 sq. ft.
5. 102 sq. ft.

Answer : Option A

Explanation :

Let l = 9 ft.

Then l + 2b = 37
=> 2b = 37 – l = 37 – 9 = 28
=> b = 282 = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.

 

2. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?
3. 400
4. 365
5. 385
6. 315

Answer : Option D

Explanation :

Let the areas of the parts be x hectares and (700 – x) hectares.

Difference of the areas of the two parts = x – (700 – x) = 2x – 700

one-fifth of the average of the two areas = 15[x+(700−x)]2
=15×7002=3505=70

Given that difference of the areas of the two parts = one-fifth of the average of the two areas
=> 2x – 700 = 70
=> 2x = 770
⇒x=7702=385

Hence, area of smaller part = (700 – x) = (700 – 385) = 315 hectares.

 

3. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle?
4. 18 cm
5. 16 cm
6. 40 cm
7. 20 cm

Answer : Option C

Explanation :

Let breadth = x cm
Then length = 2x cm
Area = lb = x × 2x = 2x²

New length = (2x – 5)
New breadth = (x + 5)
New Area = lb = (2x – 5)(x + 5)

But given that new area = initial area + 75 sq.cm.
=> (2x – 5)(x + 5) = 2x² + 75
=> 2x² + 10x – 5x – 25 = 2x² + 75
=> 5x – 25 = 75
=> 5x = 75 + 25 = 100
=> x = 1005 = 20 cm

Length = 2x = 2 × 20 = 40cm

 

4. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)?
5. 142000
6. 112800
7. 142500
8. 153600

Answer : Option D

Explanation :

l : b = 3 : 2 —-(Equation 1)

Perimeter of the rectangular park
= Distance travelled by the man at the speed of 12 km/hr in 8 minutes
= speed × time = 12×860     (∵ 8 minute = 860 hour)
= 85 km = 85 × 1000 m = 1600 m

Perimeter = 2(l + b)

=> 2(l + b) = 1600
=> l + b = 16002 = 800 m —-(Equation 2)

From (Equation 1) and (Equation 2)
l = 800 × 35 = 480 m
b = 800 × 25 = 320 m (Or b = 800 – 480 = 320m)

Area = lb = 480 × 320 = 153600 m²

 

5. It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 per square metre., what will be the total cost of the construction?
6. Rs.3500
7. Rs. 4200
8. Insufficient Data
9. Rs. 4400

Answer : Option C

Explanation :
Let length and width of the rectangular plot be l and b respectively
Total area of the rectangular plot = 96 sq.m.
=> lb = 96

Width of the pathway = 2 m
Length of the remaining area in the plot = (l – 4)
breadth of the remaining area in the plot = (b – 4)
Area of the remaining area in the plot = (l – 4)(b – 4)

Area of the pathway
= Total area of the rectangular plot – remaining area in the plot
= 96 – [(l – 4)(b – 4)] = 96 – [lb – 4l – 4b + 16] = 96 – [96 – 4l – 4b + 16] = 96 – 96 + 4l + 4b – 16
= 4l + 4b – 16
= 4(l + b) – 16

We do not know the values of l and b and hence area of the pathway cannot be found out. So we cannot determine total cost of the construction.

 

6. A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle?
7. 144√3−48π cm²
8. 121√3−36π cm²
9. 144√3−36π cm²
10. 121√3−48π cm²

Answer : Option A

Explanation :
Area of an equilateral triangle = (3/√4)*a *a where a is length of one side of the equilateral triangle
Area of the equilateral Δ ABC = (3/√4)*a *a = (3/√4)*24*24=144√3 cm2⋯ (1)

Area of a triangle = 12bhwhere b is the base and h is the height of the triangle
Let r = radius of the inscribed circle. Then
Area of Δ ABC
= Area of Δ OBC + Area of Δ OCA + area of Δ OAB
= (½ × r × BC) + (½ × r × CA) + (½ × r × AB)
= ½ × r × (BC + CA + AB)
= ½ x r x (24 + 24 + 24)
= ½ x r x 72 = 36r cm² —-(2)

From (1) and (2),
144√3=36r⇒r=144√3/36=4√3−−−−(3)

Area of a circle = πr2 where = radius of the circle
From (3), the area of the inscribed circle = πr2=π(4√3)* (4√3)=48π⋯(4)

Hence, area of the remaining portion of the triangle
= Area of Δ ABC – Area of inscribed circle
144√3−48π cm2

 

7. What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height?
8. √11600 cm
9. √14400 cm
10. √10000 cm
11. √12040 cm

Answer : Option A

Explanation :
The longest road which can fit into the box will have one end at A and other end at G (or any other similar diagonal).
Hence the length of the longest rod = AG

Initially let’s find out AC. Consider the right angled triangle ABC

AC² = AB² + BC² = 40² + 80² = 1600 + 6400 = 8000
⇒AC = √8000 cm

Consider the right angled triangle ACG

AG² = AC² + CG²
(√8000) ²+60²=8000+3600=11600
=> AG = √11600 cm
=> Length of the longest rod = √11600cm

 

8. A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed?
9. 30
10. 44
11. 56
12. 60

Answer : Option C

Explanation :

Perimeter of a rectangle = 2(l + b)
where l is the length and b is the breadth of the rectangle

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres
Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.
Hence number of poles required = 280/5 = 56

 

This topic has many practical application in day to day life. In engineering stage it is used in surveying. The basic purpose is to find the unknown variables by observing the angle of the line of sight. This is done by using some the fact that in a right angled triangle the ratio of any two sides is a function of the angle between them. From exam point of view this is one of the more tough sections and tedious to some extent. So an aspirant must thoroughly solve all the questions given here.

Important Formulas

1. Trigonometric Basics

sinθ=oppositeside/hypotenuse=y/r

cosθ=adjacentside/hypotenuse=x/r

tanθ=oppositeside/adjacentside=y/x

cosecθ=hypotenuse/oppositeside=r/y

secθ=hypotenuse/adjacentside=r/x

cotθ=adjacentside/oppositeside=x/y

From Pythagorean theorem, x2+y2=r2 for the right angled triangle mentioned above

 

2. Basic Trigonometric Values

 

θ in degrees	θ in radians	sinθ	cosθ	tanθ
0°	0	0	1	0
30°	π/6	1/2	3/√2	1/√3
45°	π/4	1/√2	1/√2	1
60°	π/3	3/√2	1/2	√3
90°	π/2	1	0	Not defined

 

3. Trigonometric Formulas

Degrees to Radians and vice versa

360°=2π radian

 

Trigonometry – Quotient Formulas

tanθ=sinθ/cosθ

cotθ=cosθ/sinθ

 

Trigonometry – Reciprocal Formulas

cosecθ=1/sinθ

secθ=1/cosθ

cotθ=1/tanθ

 

Trigonometry – Pythagorean Formulas

sin2θ+cos2θ=1

sec2θ−tan2θ=1

cosec2θ−cot2θ=1

 

4. Angle of Elevation

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, angle of elevation is the angle between the horizontal and the line from the object to the observer’s eye (the line of sight).

i.e., angle of elevation =  AOP

5. Angle of Depression

Suppose a man from a point O looks down at an object P, placed below the level of his eye. Then, angle of depression is the angle between the horizontal and the observer’s line of sight

i.e., angle of depression =  AOP

6. Angle Bisector Theorem

Consider a triangle ABC as shown above. Let the angle bisector of angle A intersect side BC at a point D. Then BD/DC=AB/AC

(Note that an angle bisector divides the angle into two angles with equal measures.
i.e., BAD = CAD in the above diagram)

7. Few Important Values to memorize

√2=1.414, √3=1.732, √5=2.236

 

Solved Examples

Level 1

1.The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is:
A. 14.8 m	B. 6.2 m
C. 12.4 m	D. 24.8 m

 

 

Answer : Option D

Explanation :

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.

cos 60° = PQ/PR

1/2=12.4/PR

PR=2×12.4=24.8 m

2.From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is:
A. 346 m	B. 400 m
C. 312 m	D. 298 m

 

 

Answer : Option A

Explanation :

tan 30°=RQ/PQ

1/√3=200/PQ

PQ=200√3=200×1.73=346 m

3.The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
A. None of these	B. 60°
C. 45°	D. 30°

 

 

Answer : Option C

Explanation :

Consider the diagram shown above where QR represents the tree and PQ represents its shadow

We have, QR = PQ
Let QPR = θ

tan θ = QR/PQ=1 (since QR = PQ)

=> θ = 45°

i.e., required angle of elevation = 45°

4.An observer 2 m tall is 103√ m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:
A. None of these	B. 12 m
C. 14 m	D. 10 m

 

 

Answer : Option B

Explanation :

SR = PQ = 2 m

PS = QR = 10√3m

tan 30°=TS/PS

1/3=TS/10√3

TS=10√3/√3=10 m

TR = TS + SR = 10 + 2 = 12 m

5.From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower?
A. 40 m	B. 138.4 m
C. 46.24 m	D. 160 m

 

 

Answer : Option B

Explanation :

Let AC be the tower and B be the position of the bus.

Then BC = the distance of the bus from the foot of the tower.

Given that height of the tower, AC = 80 m and the angle of depression, DAB = 30°

ABC = DAB = 30° (Because DA || BC)

tan 30°=AC/BC=>tan 30°=80/BC=>BC = 80/tan 30°=80/(1/√3)=80×1.73=138.4 m

i.e., Distance of the bus from the foot of the tower = 138.4 m

6.Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3 m long?
A. 30°	B. 60°
C. 45°	D. None of these

 

 

Answer : Option B

Explanation :

Let RQ be the pole and PQ be the shadow

Given that RQ = 18 m and PQ = 6√3 m

Let the angle of elevation, RPQ = θ

From the right  PQR,

tanθ=RQ/PQ=18/6√3=3/√3=(3×√3)/( √3×√3)=3√3/3=√3

θ=tan−1(3√)=60°

 

Level 2

1.A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?
A. 8 min 17 second	B. 10 min 57 second
C. 14 min 34 second	D. 12 min 23 second

 

 

Answer : Option B

Explanation :

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car

Then, ADC = 30° , ACB = 45°

Let AB = h, BC = x, CD = y

tan 45°=AB/BC=h/x

=>1=h/x=>h=x——(1)

tan 30°=AB/BD=AB/(BC + CD)=h/(x+y)

=>1/√3=h/(x+y)

=>x + y = √3h

=>y = √3h – x

=>y = √3h−h(∵ Substituted the value of x from equation 1 )

=>y = h(√3−1)

Given that distance y is covered in 8 minutes
i.e, distance h(√3−1) is covered in 8 minutes

Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).

Let distance h is covered in t minutes

since distance is proportional to the time when the speed is constant, we have

h(√3−1)∝8—(A)

h∝t—(B)

(A)/(B)=>h(√3−1)/h=8/t

=>(√3−1)=8/t

=>t=8/(√3−1)=8/(1.73−1)=8/.73=800/73minutes ≈10 minutes 57 seconds

2.The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
A. 5 metres	B. 8 metres
C. 10 metres	D. 12 metres

 

 

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the tower and DE represents the pole

Given that AC = 15 m , ADB = 30°, AEC = 60°

Let DE = h

Then, BC = DE = h, AB = (15-h) (∵ AC=15 and BC = h), BD = CE

tan 60°=AC/CE=>√3=15/CE=>CE = 15√3— (1)

tan 30°=AB/BD=>1/√3=(15−h)/BD

=>1/√3=(15−h)/(15/√3)(∵ BD = CE and Substituted the value of CE from equation 1)

=>(15−h)=(1/√3)×(15/√3)=15/3=5

=>h=15−5=10 m

i.e., height of the electric pole = 10 m

 

3.Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 300 m	B. 173 m
C. 273 m	D. 200 m

 

 

Answer : Option C

Explanation :

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m,  BAD = 30° ,  BCD = 45°

tan 30° = BD/BA⇒1/√3=100/BA

⇒BA=100√3

tan 45° = BD/BC

⇒1=100/BC

⇒BC=100

Distance between the two ships = AC = BA + BC
=100√3+100=100(√3+1)=100(1.73+1)=100×2.73=273 m

4.From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole?
A. 52 m	B. 50 m
C. 66.67 m	D. 33.33 m

 

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the hill and DE represents the pole

Given that AC = 100 m

XAD = ADB = 30° (∵ AX || BD )
XAE = AEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE

=>√3=100/CE=>CE = 100/√3— (1)

tan 30°=AB/BD=>1/√3=(100−h)/BD

=>1/√3=(100−h)/(100/√3)(∵ BD = CE and Substituted the value of CE from equation 1 )

=>(100−h)=1/√3×100/√3=100/3=33.33=>h=100−33.33=66.67 m

i.e., the height of the pole = 66.67 m

5.A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower?
A. 9 m	B. 10.40 m
C. 15.57 m	D. 12 m

 

 

Answer : Option A

Explanation :

Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.

Given that AD = 18 m, ABC = 60°, DBC = 30°

Let DC be h.

tan 30°=DC/BC

1/√3=h/BC

h=BC√3—— (1)

tan 60°=AC/BC

√3=(18+h)/BC

18+h=BC×√3—— (2)

(1)/(2)=>h/(18+h)=(BC/√3)/(BC×√3)=1/3

=>3h=18+h=>2h=18=>h=9 m

i.e., the height of the tower = 9 m

6.A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon?
A. 0.63 meter/sec	B. 2.16 meter/sec
C. 3.87 meter/sec	D. 0.72 meter/sec

 

 

Answer : Option B

Explanation :

Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.

Given that CA = 150 m, BCA = 60°

tan 60°=BA/CA

√3=BA/150

BA=150√3

i.e, the distance travelled by the balloon = 150√3meters

time taken = 2 min = 2 × 60 = 120 seconds

Speed = Distance/Time=150√3/120=1.25√3=1.25×1.73=2.16 meter/second

7. The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree?
A. 22 m	B. 44 m
C. 33 m	D. None of these

 

 

Answer : Option B

Explanation :

Let DC be the wall, AB be the tree.

Given that DBC = 30°, DAE = 60°, DC = 11 m

tan 30°=DC/BC

1/√3=11/BC

BC = 11√3 m

AE = BC =11√3 m—— (1)

tan 60°=ED/AE

√3=ED/11√3[∵ Substituted the value of AE from (1)]

ED =11√3×√3=11×3=33

Height of the tree = AB = EC = (ED + DC) = (33 + 11) = 44 m

 

8. Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles.
A. 141 m and 282 m	B. 70.5 m and 141 m
C. 65 m and 130 m	D. 130 m and 260 m

 

 

Answer : Option B

Explanation :

Let AB and CD be the poles with heights h and 2h respectively

Given that distance between the poles, BD = 200 m

Let E be the middle point of BD.

Let AEB = θ and CED = (90-θ) (∵ given that angular elevations are complementary)

Since E is the middle point of BD, we have BE = ED = 100 m

From the right  ABE,
tanθ=AB/BE and tanθ=h/100

h = 100tanθ—— (1)

From the right  EDC,

tan(90−θ)=CD/ED

cotθ=2h/100[∵tan(90−θ)=cotθ]

2h =100cotθ—— (2)

(1) × (2) => 2h2=1002[∵tanθ×cotθ=tanθ×1/tanθ=1]

=>√2h=100

=>h=100/√2=(100×√2)/( √2×√2)=50√2=50×1.41=70.5

2h=2×70.5=141

i.e., the height of the poles are 70.5 m and 141 m.

9. To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window?
A. 8.65 m	B. 2 m
C. 2.5 m	D. 3.65 m

 

 

Answer : Option D

Explanation :

Let AB be the man and CD be the window

Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
DAF = 45° , CAF = 60°

From the diagram, AF = BE = 5 m

From the right  AFD, tan45°=DF/AF

1=DF/5

DF = 5—— (1)From the right  AFC, tan60°=CF/AF

√3=CF/5

CF=5√3—— (2)

Length of the window = CD = (CF – DF)

=5√3−5[∵ Substitued the value of CF and DF from (1) and (2)]=5(√3−1)=5(1.73−1)=5×0.73=3.65 m

10.The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°, the elevation changes to 60°. What is the approximate height of the mountain?
A. 1.2 km	B. 0.6 km
C. 1.4 km	D. 2.7 km

 

 

Answer : Option D

Explanation :

Let A be the foot and C be the summit of a mountain.

Given that CAB = 45°

From the diagram, CB is the height of the mountain. Let CB = x

Let D be the point after ascending 2 km towards the mountain such that
AD = 2 km and given that DAY = 30°

It is also given that from the point D, the elevation is 60°

i.e., CDE = 60°

From the right  ABC,

tan45°=CB/AB

=>1=x/AB[∵ CB = x (the height of the mountain)]

=>AB = x—— (eq:1)

From the right  AYD,

sin30°=DY/AD

=>1/2=DY/2(∵ Given that AD = 2)

=> DY=1—— (eq:2)

cos30°=AY/AD=>√3/2=AY/2(∵ Given that AD = 2)=> AY=√3—— (eq:3)

From the right  CED, tan60°=CE/DE=>tan60°=(CB – EB)/YB∵ [CE = (CB – EB) and DE = YB)]

=>tan60°=(CB – DY)/(AB – AY)[ ∵ EB = DY and YB = (AB – AY)]

=>tan60°=(x – 1)/(x -√3)∵ [CB = x, DY = 1(eq:2), AB=x (eq:1) and AY = 3√(eq:3)]

=>√3=(x – 1)/(x -√3)=>x√3−3=x−1=>x(√3−1)=2=>0.73x=2=>x=2/0.73=2.7

i.e., the height of the mountain = 2.7 km

 

Classification involves putting things into a class or group according to particular characteristics so it’s easier to make sense of them, whether you’re organizing your shoes, your stock portfolio, or a group of invertebrates.  From all competitive examination classification is one of the most important topics, this pattern come with lot of questions minimum they asking the 4 to 5 question from the classification topic. In the SSC CGL or SSC constable GD examination having the same topics from the reasoning section but the standard of the topic will be different, so most of the candidates preference for this topic to get the best score in the written examination.

 

 

Directions: Find the odd one out

 

1. A. Square B. Circle                     C. Rectangle             D. Triangle

 

2. A. Cotton B. Terene                  C. Silk                         D. Wool

 

3. A. Light B. Wave                    C. Heat                      D. Sound

 

4. A. 81 : 243 B. 16 :64                   C. 64 : 192                D. 25 : 75

 

5. A. 64 : 8 B. 80 : 9                     C. 7 : 49                     D. 36 : 6

 

6. A. 26 : 62 B. 36 : 63                  C. 46 : 64                  D. 56 : 18

 

7. A. ABZY B. BCYX                      C. CDVW                   D. DEVU

 

8. A. ACE B. FHJ                         C. KLM                       D. SUW

 

9. Find the wrong number in the series

441, 484, 529, 566, 625

1. 484 B. 529                                    C. 625                                    D. 566

 

10. Find the wrong number in the series

232, 343, 454, 564, 676

1. 676 B. 454                                    C. 343                                    D. 564

 

 

SOLUTION TO CLASSIFICATION LEVEL 1

 

 

1. B. Except circle, all others are geometrical figures consisting straight lines.

 

2. B. Except terene, all others are natural fibres.

 

3. B. Except wave, all others are different form of energy.

 

4. B. 81*3=243

64*3=192

25*3=75

But     16*4=64

 

5. D. Except D, in each pair one number is square root of the other.

 

6. D. Except D, in each pair the position of digits has been interchanged.

 

7. C. A+1=B   &   Z-1=Y

B+1=C   &   Y-1=X

D+1=E   &   V-1=U

But   C+1=D   &   V+1=W

 

8. C. A+2=C    &   C+2=E

F+2=H     &   H+2=J

But      K+1=L     &   L+1=M

 

9. D. 21^2=441

22^2=484

23^2=529

25^2=625

But   (23.79)^2=566

 

10. D. 232+111=343

343+111=454

454+111=565 (but given 564)

 

This module will teach you the basics of direct and indirect proportions. These concepts will further help you in time and work questions.

Important Formulas – chain rule

• Direct Proportion

  Two quantities are said to be directly proportional, if on the increase or decrease of the one, the other increases or decreases the same extent.
  Examples

  1. Cost of the goods is directly proportional to the number of goods. (More goods, More cost)
  2. Amount of work done is directly proportional to the number of persons who did the work. (More persons, More Work)

• Indirect Proportion (inverse proportion)

  Two quantities are said to be indirectly proportional (inversely proportional) if on the increase of the one, the other decreases to the same extent and vice-versa.

Examples

• 1. Number of days needed to complete a work is indirectly proportional (inversely proportional) with the number of persons who does the work (More Persons, Less Days needed)
  2. The time taken to travel a distance is indirectly proportional (inversely proportional) with the speed in which one is travelling (More Speed, Less Time)

 

Solved Examples

Level 1

1. If the cost of x metres of wire is d rupees, then what is the cost of y metres of wire at the same rate?
A. Rs. (xd/y)	B. Rs. x/d
C. Rs. (yd/x)	D. Rs. y/d

 

Answer : Option C

Explanation :

cost of x metres of wire = Rs. d

cost of 1 metre of wire = Rs.(d/x)

cost of y metre of wire = Rs.(y×d/x)=Rs. (yd/x)

2. In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?
A. 50	B. 30
C. 40	D. 10

 

Answer : Option B

Explanation :

Meal for 200 children = Meal for 120 men

Meal for 1 child = Meal for 120/200 men

Meal for 150 children = Meal for (120×150)/200 men=Meal for 90 men

Total mean available = Meal for 120 men

Renaming meal = Meal for 120 men – Meal for 90 men = Meal for 30 men

 

3. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work?
A. 26	B. 22
C. 12	D. 24

 

Answer : Option D

Explanation :
Let the required number of days be x

More men, less days (indirect proportion)

Hence we can write as

Men36:27}::x:18 ⇒36×18=27×x ⇒12×18=9×x

⇒12×2=x

⇒x=24

4. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made 21 revolutions, what will be the number of revolutions made by the larger wheel?
A. 15	B. 12
C. 21	D. 9

 

Answer : Option D

Explanation :

Let the number of revolutions made by the larger wheel be x

More cogs, less revolutions (Indirect proportion)

Hence we can write as

Cogs 6:14}: x: 21⇒6×21=14×x ⇒6×3=2×x ⇒3×3=x ⇒x=9

5. 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day should 4 pumps work in order to empty the tank in 1 day?
A. 10	B. 12
C. 8	D. 15

 

Answer : Option B

Explanation :

Let the required hours needed be x

More pumps, less hours (Indirect proportion)
More Days, less hours (Indirect proportion)

Hence we can write as

Pumps  3:4

::x:8

Days                      2:1

⇒3×2×8=4×1×x

⇒3×2×2=x

⇒x=12

6. 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work?
A. 9	B. 12
C. 10	D. 13

 

Answer : Option D

Explanation :
Let the required number of days be x

More persons, less days (indirect proportion)
More hours, less days (indirect proportion)

Hence we can write as

Persons                39:30

::x:12

Hours    5:6
⇒39×5×12=30×6×x ⇒39×5×2=30×x ⇒39=3×x ⇒x=13

7. A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how many seconds will it take for the loom to weave 25 meters of cloth?
A. 205	B. 200
C. 180	D. 195

 

Answer : Option D

Explanation :

Let the required number of seconds be x

More cloth, More time, (direct proportion)

Hence we can write as

Cloth         0.128:25} :: 1:x

⇒0.128x=25 ⇒x=25/0.128 ⇒25000/128=3125/16≈195

 

8. 21 goats eat as much as 15 cows. How many goats each as much as 35 cows?
A. 49	B. 32
C. 36	D. 41

 

Answer : Option A

Explanation :

15 cows ≡ 21 goats

1 cow ≡21/15 goats

35 cows ≡ (21×35)/15 goats≡(21×7)/3 goats≡7×7 goats ≡ 49 goats

 

Level 2

 

1. In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?
A. 1	B. 40
C. 20	D. 26

 

Answer : Option B

Explanation :

Assume that in x days, one cow will eat one bag of husk.

More cows, less days (Indirect proportion)
More bags, more days (direct proportion)
Hence we can write as

Cows    40:1         ::x:40

Bags     1:40

⇒40×1×40=1×40×x ⇒x=40

2. If a quarter kg of potato costs 60 paise, how many paise does 200 gm cost?
A. 65 paise	B. 70 paise
C. 52 paise	D. 48 paise

 

Answer : Option D

Explanation :
Let 200 gm potato costs x paise

Cost of ¼ Kg potato = 60 Paise
=> Cost of 250 gm potato = 60 Paise (∵ 1 Kg = 1000 gm => ¼ Kg = ¹⁰⁰⁰/₄ gm = 250 gm)

More quantity, More Paise (direct proportion)

Hence we can write as

Quantity  200:250} :: x:60

⇒200×60=250×x ⇒4×60=5×x ⇒4×12=x ⇒x=48

3. A contract is to be completed in 56 days if 104 persons work, each working at 8 hours a day. After 30 days, 2/5 of the work is completed. How many additional persons should be deployed so that the work will be completed in the scheduled time, each person’s now working 9 hours a day.
A. 160	B. 150
C. 24	D. 56

 

Answer : Option D

Explanation :

Persons worked = 104
Number of hours each person worked per day = 8
Number of days they worked = 30
Work completed = ²/₅

Remaining days = 56 – 30 = 26
Remaining Work to be completed = 1 – ²/₅ = ³/₅
Let the total number of persons who do the remaining work = x
Number of hours each person needs to be work per day = 9

More days, less persons(indirect proportion) More hours, less persons(indirect proportion)
More work, more persons(direct proportion)

Hence we can write as

Days     30:26

Hours    8:9                                   ::x:104

Work     35:25
⇒30×8×3/5×104=26×9×2/5×x

⇒x=(30×8×3/5×104)/(26×9×2/5)=(30×8×3×104)/(26×9×2)

=(30×8×104)/(26×3×2)=(30×8×4)/(3×2)=5×8×4=160

Number of additional persons required = 160 – 104 = 56

 

4. x men working x hours per day can do x units of a work in x days. How much work can be completed by y men working y hours per day in y days?
A. x2/y2 units	B. y3/x2 units
C. x3/y2 units	D. y2/x2 units

 

Answer : Option B

Explanation :
Let amount of work completed by y men working y hours per in y days = w units

More men, more work(direct proportion)
More hours, more work(direct proportion)
More days, more work(direct proportion)

Hence we can write as

Men                      x:y

Hours    x:y          ::x:w

Days                      x:y
⇒x³w=y³x ⇒w=y³x/x³=y³/x²

5. A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building, which casts a shadow of length 28.75 m under similar conditions?
A. 12.5 m	B. 10.5 m
C. 14	D. 12

 

Answer : Option A

Explanation :
Let the required height of the building be x meter

More shadow length, More height (direct proportion)

Hence we can write as

Shadow length 40.25:28.75}:: 17.5:x

⇒40.25×x=28.75×17.5 ⇒x=(28.75×17.5)/40.25=(2875×175)/40250

= (2875×7)/1610=2875/230=575/46=12.5

 

6. If the price of 357 apples is Rs.1517.25, what will be the approximate price of 49 dozens of such apples?
A. Rs. 2500	B. Rs. 2300
C. Rs. 2200	D. Rs. 1400

 

Answer : Option A

Explanation :

Let the required price be x

More apples, More price (direct proportion)

Hence we can write as

Apples 357:(49×12)} :: 1517.25:x

⇒357x = (49×12)×1517.25⇒x = (49×12×1517.25)/357=(7×12×1517.25)/51

= (7×4×1517.25)/17

=7×4×89.25≈2500

7. 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal is required for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as 4 engines of latter type?
A. 20 metric tonnes	B. 22 metric tonnes
C. 24 metric tonnes	D. 26 metric tonnes

 

Answer : Option D

Explanation :

Let required amount of coal be x metric tonnes

More engines, more amount of coal (direct proportion)

If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption.
If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption
More rate of consumption, more amount of coal (direct proportion)

More hours, more amount of coal(direct proportion)

Hence we can write as

Engines                                                                9:8

rate of consumption                       13:14                     ::24:x

hours                                                                    8:13
⇒9×1/3×8×x=8×1/4×13×24 ⇒3×8×x=8×6×13 ⇒3xX=6×13⇒x=2×13=26

8. in a camp, food was was sufficient for 2000 people for 54 days. After 15 days, more people came and the food last only for 20 more days. How many people came?
A. 1900	B. 1800
C. 1940	D. 2000

 

Answer : Option A

Explanation :

Given that food was sufficient for 2000 people for 54 days
Hence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (∵ 54 – 15 =39)
Let x number of people came after 15 days.
Then, total number of people after 15 days = (2000 + x)
Then, the remaining food was sufficient for (2000 + x) people for 20 days

More men, Less days (Indirect Proportion)⇒Men        2000:(2000+x)}  ::  20:39

⇒2000×39=(2000+x)20⇒100×39=(2000+x)⇒3900=2000+x⇒x=3900−2000=1900